Using the balanced equation:3 Ag2S + 2 Al -> Al2S3 + 6 Ag0.88 moles of Ag2S = ? grams Ag

Answers

Answer 1

Answer: 189.5g of Ag would be produced from 0.88 moles of Ag2S

Explanation:

The question requires us the amount of Ag, in grams, that would be produced from 0.88 moles of Ag2S, according to the following equation:

[tex]3Ag_2S+2Al\rightarrow Al_2S_3+6Ag[/tex]

To solve this problem, which is a mol to mass stoichiometry problem, we'll need to follow the steps:

moles of Ag2S → moles of Ag → mass of Ag

To do that, we'll need to use the molar mass of Ag (107.87 g/mol) and the mole ratio between Ag2S and Ag, as given by the balanced chemical equation (6 moles of Ag are produced from 3 moles of Ag2S).

We can calculate the number of moles of Ag that would be produced from 0.88 moles of Ag2S as:

3 mol Ag2S -------------------------- 6 mol Ag

0.88 mol Ag2S --------------------- x

Solving for x, we'll have:

[tex]x=0.88mol\text{ Ag}_2S\times\frac{6mol\text{ Ag}}{3mol\text{ Ag}_2S}=1.76mol\text{ Ag}[/tex]

Therefore, 1.76 moles of Ag would be produced from 0.88 moles of Ag2S.

Next, we need to convert the number of moles of Ag calculated to its correspondent mass. Knowing that the molar mass of Ag is 107.87 g/mol, we can calculate:

[tex]\begin{gathered} number\text{ of moles = }\frac{mass\text{ \lparen g\rparen}}{molar\text{ mass \lparen g/mol\rparen}}\rightarrow mass\text{ = number of moles}\times molar\text{ mass} \\ \\ mass\text{ = 1.76mol}\times107.87g/mol=189.8g \end{gathered}[/tex]

Therefore, 189.5g of Ag would be produced from 0.88 moles of Ag2S.


Related Questions

This type of bond is between two metals described as a "Sea of Electrons'?

Answers

Answer:

In the early 1900's, Paul Drüde came up with the "sea of electrons" metallic bonding

The correct answer would be Metallic Bond

Find the element that is oxidized and the one that is reduced Si + 2 F2 --> SiF4

Answers

Answer

The element that is oxidized is Si and the one that is reduced is F₂.

Explanation

Si + 2F₂ → SiF₄

The given reaction is an oxidation-reduction (redox) reaction:

Oxidation: Si → Si⁴⁺ + 4e⁻

Reduction: 2F₂ + 4e⁻ → 4F⁻

Si is a reducing agent, and F₂ is an oxidizing agent.

An oxidizing agent gains electrons and is reduced in a chemical reaction.

A reducing agent loses electrons and is oxidized in a chemical reaction.

Therefore, the element that is oxidized is Si and the one that is reduced is F₂.

Lithium nitride is an ionic compound. Draw a diagram which shows its formula, the charges on the ions and the arrangement of the valency (outer shell) electrons around the negative ion. Use x for an electron from a lithium atom and o for an electron from a nitrogen atom. ​

Answers

Lithium nitride is an ionic compound and the chemical formula of lithium nitride is Li₃N and the charges on the ions is Li⁺¹N³⁻ 3 valence electron in lithium and 5 valence electron in nitrogen

Ionic compound is the ion compound these ions are atom that gain or lose electron resulting in a net positive and negative charges and lithium nitride is an ionic compound and in that Li⁺¹N³⁻ is the ion and the arrangement of the valency (outer shell) electrons are 3 valence electron in lithium and 5 valence electron in nitrogen

X = 3 electron

O = 7 electron

And that's why lithium nitride has 10 electron

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an equimolar mixture of n2(g)n2(g) and ar(g)ar(g) is kept inside a rigid container at a constant temperature of 300 kk. the initial partial pressure of arar in the mixture is 0.75atm0.75atm. an additional amount of arar was added to the container, enough to double the number of moles of arar gas in the mixture. assuming ideal behavior, what is the final pressure of the gas mixture after the addition of the arar gas? responses 0.75atm0.75atm, because increasing the partial pressure of arar decreases the partial pressure of n2n2. 0.75 atmosphere , because increasing the partial pressure of a r decreases the partial pressure of n 2 . 1.13atm1.13atm, because 333% of the moles of gas are n2n2. 1.13 atmospheres , because 33 percent of the moles of gas are n 2 . 1.50atm1.50atm, because the number of moles of n2n2 did not change. 1.50 atmospheres , because the number of moles of n 2 did not change. 2.25atm2.25atm, because doubling the number of moles of arar doubles its partial pressure.

Answers

To solve such type of question we must be knowing the concept behind the ideal gas equation. The final pressure of the gas mixture after the addition of the Ar gas is 2.25 atm

What is ideal gas equation?

Ideal gas equation is the mathematical expression that relates pressure volume, temperature and number of moles of gas

Mathematically,

PV=nRT

according to question T and V is constant

P ∝ n

P₁/n₁= P₂/n₂

Where

P₁ = initial pressure= 0.75atm

P₂ = Final pressure=?

n₁= number of moles of gas initially present=n

n₂ = Final moles of gas present=2n

Substituting into the given equation

P₂= P₁n₂/n₁

P₂ = 0.75atm ×2n/n

P₂ = 1.5 atm

The total pressure of the gas=partial pressure of N2 +  partial pressure after addition of Ar = 0.75 atm + 1.5 atm = 2.25 atm

The final pressure of the gas mixture after the addition of the Ar gas is 2.25 atm

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At 302 K , to what pressure can the carbon dioxide in the cartridge inflate a 3.05 L mountain bike tire? (Note that the gauge pressure is the difference between the total pressure and atmospheric pressure. In this case, assume that atmospheric pressure is 14.7 psi .)

Answers

Ideal gas law is valid only for ideal gas not for vanderwaal gas. Therefore the carbon dioxide in the cartridge inflate a 3.05 L  mountain bike tire to 104.3psi pressure.

What is ideal gas equation?

Ideal gas equation is the mathematical expression that relates pressure volume and temperature.

Mathematically,

PV=nRT

where,

P = pressure

V= volume=3.05 L

n =number of moles=1mole(assumed as it is not given in question)

T =temperature =  302 K

R = Gas constant = 0.0821 L.atm/K.mol

P × 3.05 L  =1 mole× 0.0821 L.atm/K.mol ×  302 K

P =8.12atm

1 atm = 14.7 psi

Hence Pressure in psi  = 8.12atm×14.7 psi = 119.49 psi

Pressure by the gas= Total pressure - Atmospheric pressure = 119.49 - 14.7 psi = 104.3psi

Therefore the carbon dioxide in the cartridge inflate a 3.79 L mountain bike tire to 104.3psi pressure.

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Name three different forms of mixture

Answers

Answer:

Mixtures can be classified on the basis of particle size into three different types: solutions, suspensions and colloids. The components of a mixture retain their own physical properties.

Sorry for the bad English, love from Vanuatu!

Explanation:

What is the balanced chemical equation for the reaction of sodium and oxygen?Na + O to NaO2Na + O2 to 2 NaO4Na + O2 to 2Na2ONa + O2 to NaO2

Answers

Answer

4Na + O₂ → 2Na₂O

Explanation

Sodium reacts with oxygen to form sodium oxide and has the following balanced chemical equation:

4Na + O₂ → 2Na₂O.

Hence, the balanced chemical equation for the reaction of sodium and oxygen is:

4Na + O2 to 2Na2O

Which of the following does NOT happen during a chemical reaction?A. Bonds are brokenB. Bonds are formedC. Energy is createdD. Mass is conserved

Answers

ANSWER

Energy is created ------ (option C)

EXPLANATION

During the chemical reaction process, reactants need to be converted to form products. To balance such chemical reaction equation, we need to apply the law of conservation of mass

Law of conservation of mass states that matter can neither be created nor destroyed but can be transformed from one form to another.

In the process ofa chemical reaction, bonds are broken are the reactants side, bonds are formed on the products side and mass is conserved

Therefore, the correct answer is Energy is created

17. Which of the following represents a formula for a chemical compound?A. CB. KOHC. O

Answers

Answer:

KOH. Option B is correct

Explanations:

A chemical compound are made up of more than one element combined together. According to the question, we need to determine the formula that represents a compound.

The compound there is KOH since it contains three elements (Potassium, Oxygen and Hydrogen)

a) A solution of sodium thiosulfate, Na2S2O3, is 0.1047 M. 18.59 mL of this solution reacts with 38.62 mL of I2 solution. What is the molarity of the I2 solution?

2(Na2S2O3) + I2↔Na2S4O6 + 2(NaI)

b) 26.64 mL of the I2 solution from above is required to titrate a sample containing As2O3. Calculate the mass of As2O3 (197.8 g/mol) in the sample.

As2O3 + 5(H2O) + 2I2 → 2(H3AsO4) + 4HI

Answers

From the calculations, the molarity of the iodine solution is 0.025 M and the mass of the arsenic oxide solution is 0.066 g.

What is the molarity?

Let us recall that the molarity is defined as the ratio of the number of moles to the volume of the solution. It is a measure of the amount of substance present. Ley us now try to use what we know to obtain the molarity of the solution in each of the cases of the questions.

a) Using;

CAVA/CBVB = NA/NB

CA = Amount of thiosuphate

CB = Amount of iodine

VA = volume of thiosuphate

VB = volume of iodine

NA = number of moles of thisulphate

NB = Number of moles of iodine

Hence;

0.1047 M * 18.59 mL/CB * 38.62 mL = 2/1

0.1047 M * 18.59 mL * 1 = CB * 38.62 mL * 2

CB = 0.1047 M * 18.59 mL * 1 /38.62 mL * 2

CB = 1.946/77.2

CB = 0.025 M

b) Using the formula;

Number of moles = concentration * volume

Number of moles = 0.025 M * 26.64 /1000 L

= 6.66 * 10^-4 moles

Again;

1 mole of the arsenic oxide reacts with 2 moles of iodine solution

x moles of the arsenic oxide reacts with 6.66 * 10^-4 moles of iodine

x = 1 mole *  6.66 * 10^-4 moles/ 2 moles

x = 3.33  * 10^-4 moles

Mass of the arsenic oxide = 3.33  * 10^-4 moles * 197.8 g/mol

= 0.066 g

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The value of AH rxn for the following reaction is -72 kJ. How many kJ of heat is released when 0.989 g of HBr (80.91 g/mol) is formed? H2 (g) + Br2 (g) -› 2 HBr (gram). A. -144B. -72 C. -0.44 D. -36

Answers

Answer:

[tex]C\text{ : -0.44 KJ}[/tex]

Explanation:

Here, we want to get the amount of heat released in KJ

From the change in enthalpy given and the equation of reaction, we know that 2 moles of HBr would lead to that amount of heat

Now, let us get the actual amount of heat released

We need to get the actual number of moles of HBr produced

Mathematically, we can calculate that by dividing the mass of HBr by its molar mass

We have that as:

[tex]\frac{0.989}{80.91}\text{ = 0.0122 mol}[/tex]

From the reaction information:

-72 KJ was released by 2 moles

x KJ would be released by 0.0122 mol

To get the value of x, we have it that:

[tex]\begin{gathered} x\text{ }\times2\text{ = 0.0122 }\times\text{ \lparen-72\rparen} \\ \\ x\text{ = -36 }\times\text{ \lparen0.0122\rparen = -0.44 KJ} \end{gathered}[/tex]

How many atoms of O are there in 7.00 g FeSO4 ?

Answers

Answer:

1.11 x 10²³ atoms O

Explanation:

To find the number of oxygen atoms in FeSO₄, you need to (1) convert FeSO₄ from grams to moles (using the molar mass), then (2) convert moles FeSO₄ to moles O (using the mole-to-mole ratio of FeSO₄), and then (3) convert moles O to atoms O (using Avogadro's Number). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs like the given number (7.00 = 3 sig figs).

Atomic Mass (Fe): 55.845 g/mol

Atomic Mass (S): 32.065 g/mol

Atomic Mass (O): 15.999 g/mol

Molar Mass (FeSO₄): 55.845 g/mol + 32.065 g/mol + 4(15.999 g/mol)

Molar Mass (FeSO₄): 151.906 g/mol

1 mole FeSO₄: 1 mole Fe, 2 mole S, 4 moles O

Avogadro's Number:

6.022 x 10²³ atoms = 1 mole

 7.00 g FeSO₄             1 mole                 4 moles O            6.022 x 10²³ atoms
-----------------------  x  -------------------  x  -----------------------  x  ----------------------------
                                   151.906 g           1 mole FeSO₄                  1 mole

=  1.11 x 10²³ atoms O

One day in lab, while adding a gnarled root to a dark liquid bubbling in an iron cauldron, your friend Leila (an expert chemist) says this:
"Group 1A metal hydrides react with water to produce hydroxides and hydrogen gas."
Using Leila's statement, and what you already know about chemistry, predict the products of the following reaction.

Be sure your chemical equation is balanced!

KH(s)+ H2O(l)=

Answers

The balanced equation of the reaction of the metal hydride, KH, and water is given below:

KH (s) + H₂O(l) ---> KOH + H₂ (g)

What are metal hydrides?

Metal hydrides are compounds that are composed of a reactive metal chemically combined with hydrogen.

The hydrogen atom is less electropositive than the metal and will accept electrons from the metal to form the negative hydride ion.

For example, Group 1A metal hydrides react with water to produce hydroxides and hydrogen gas.

Considering the above property, the given reaction will be:

KH (s) + H₂O(l) ---> KOH + H₂ (g)

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Calculate the total energy of 9.4 * 10^16 photons of visible light with a wavelength of 4.3 * 10^7 m.

Answers

According to the Planck's equation which is E= hc/λ the total energy  of 9.4×10¹⁶ photons is 43.446×10[tex]^-17[/tex] J.

What is Planck's equation?

Max Planck  discovered the theory which stated that energy is transferred in the form of discrete packs which are called quanta  and thus proposed an equation called the Planck's equation which relates energy and frequency of a photon and is given as, E=hcυ or in terms of wavelength it is ,E=hc/λ.

The equation makes use of a constant which is called the Planck's constant and it's value is 6.626×10[tex]^-34[/tex] Js.

Substituting the given value of wavelength of  one photon  in the above formula containing wavelength,E=6.626×10[tex]^-34[/tex]×3×10⁸/4.3×10⁷=4.622×10[tex]^-33[/tex] J.

Now, for energy of 9.4×10¹⁶ photons =4.622×10[tex]^-33[/tex]×9.4×10¹⁶=43.446×10[tex]^-17[/tex] J.

Thus, the energy of 9.4×10¹⁶ photons is 43.446×10[tex]-17[/tex] J.

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calculate the theoretical and % yield of copper when 0.500 g of Cu was used and 0.350 g were recovered at the end of the experiment. complete solution

Answers

The theoretical percentage yield of copper  is 70%

This is further explained below.

What is a yield?

Generally, the equation for  yield is mathematically given as

[tex]yield =\frac{\text { mote no. of product }}{\text { mote no. of reactant }} \times 100$[/tex]

Here, Based on Theoretical Yield

[tex]\begin{aligned} \text { mole rio. of reactant } &=-\frac{0.5}{63.546} \text { mole } \\ \end{aligned}$[/tex]

Based on the Actual yield

mole rio. of Product=0.35/63.546 mole

(Here atomic man of Cu=63.546g/mol )

Therefore, we apply the initially stated equation for %yield

So,  %yield =[tex]\frac{\frac{0.35}{63.546}}{\frac{0.5}{(63.546}} \times 100 \%$[/tex]

=[tex]\frac{0.35}{0.5} \times 100 \%\\[/tex]

=0.7 *100%

=70%

In conclusion,  the percentage yield is 70%

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Which of the following is the ground-state electron configuration of a calcium atom?
O a. [Ne]3s²
O b. [Ne]3s²3p6
OC. [Ar]3d²
O d. [Ar]4s²
O e. [Ar]4s¹3d1

Answers

The arrangement of electrons in orbitals around an atomic nucleus is known as electronic configuration. It is also known as electronic structure or electron configuration. It follows Aufbau's principle, Hund's rule and Pauli's exclusion principle, to fill the electrons in various orbitals according to the energy of the orbitals.

Calcium has 20 electrons, that is, the atomic number of Ca = 20

Hence its electronic configuration will be, 1[tex]s^{2}[/tex] 2[tex]s^{2}[/tex] 2[tex]p^{6}[/tex] 3[tex]s^{2}[/tex] 3[tex]p^{6}[/tex] 4[tex]s^{2}[/tex]

Now, since Argon has 18 electrons, that is, the atomic number of Ar = 18, its electronic configuration will be,

                             Ar =   1[tex]s^{2}[/tex] 2[tex]s^{2}[/tex] 2[tex]p^{6}[/tex] 3[tex]s^{2}[/tex] 3[tex]p^{6}[/tex]

Thus , the simplified electronic configuration of Calcium is,

                                 Ca = [Ar]4[tex]s^{2}[/tex]

Thus, option d)  [Ar]4[tex]s^{2}[/tex] is the correct answer.

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Usually scientists will a large isotope such as Uranium-235 with a ____.
The collision results in several more nuclear fissions, also known as a _____. This results in a HUGE release in energy!

Answers

Usually, scientists will collide a large isotope such as Uranium-235 with a neutron. The collision results in several more nuclear fissions, also known as a chain reaction. This results in a HUGE release of energy!

What is nuclear fission?

Nuclear fission is a type of nuclear reaction in which a large isotope such as uranium-235b is split into two or more smaller isotopes with the release of large quantities of energy and radiation.

The process of nuclear fission in large isotopes can be initiated by colliding or bombarding these large isotopes with fast-moving neutrons which splits the nucleus of the atom, releasing energy ad more neutrons that will split more atoms of the large isotope, resulting in a chain reaction.

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what is the volume (in liters) of a 4.2 gram sample of O2 at STP

Answers

to solve this we need to use the ideal gas law:

[tex]PV=nRT[/tex]

we need to calculate the number of mols of O2

[tex]n=\frac{m}{Mm}=\frac{4.2g}{32\text{ g/mol}}=0.131mol[/tex]

Now we just have to solve for volume and use the number of mols just calculated and the standard pressure (1 atm) and temperature 273.15K:

[tex]V=\frac{nRT}{P}=\frac{0.131mol\cdot0.082\text{ }\frac{atm\cdot l}{K\cdot mol}273.15K}{1atm}=2.94l[/tex]

List two components of rocket fuel?

Answers

A component that many space agencies use is liquid hydrogen, which can be abbreviated as LH2.

And another fuel very used in rockets is kerosene which is a hydrocarbons mixture of other compounds that fuel.

Answer:

Most liquid chemical rockets use two separate propellants: a fuel and an oxidizer. Typical fuels include kerosene, alcohol, hydrazine and its derivatives, and liquid hydrogen. Many others have been tested and used. Oxidizers include nitric acid, nitrogen tetroxide, liquid oxygen, and liquid fluorine.

Explanation:

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3. If you need to produce 85 g of CO2, how many grams of: (these are 3 problems starting with thea. C3H8, do you need?same amount:b. O2, do you need?c. H2O will also be made?

Answers

1) First let's write the equation. It is a combustion reaction, so:

C₃H₈ + O₂ ---> CO₂ + H₂O

and balance the equation (same number of atoms of each element on both sides of the equation):

C₃H₈ + 5 O₂ ---> 3 CO₂ + 4 H₂O

Reactant side:

C - 3

H - 8

O - 10

Product side:

C - 3

O - 10

H - 8

2) Now let's transform 85 grams of CO₂ into mole. For this, we use the following equation:

mole = mass/molar mass

Molar mass of CO₂ is: (1×12) + (2×16) = 44 g/mol

mole = 85/44

mole = 1.9 mol of CO₂

3) Now we use the proportion of the balanced equation:

1 mol of C₃H₈ ---- 3 mol of CO₂

x mol of C₃H₈ ----- 1.9 mol of CO₂

x = 0.6 mol of C₃H₈

4) Now we transform mole of C₃H₈ into grams using its molar mass.

molar mass of C₃H₈ is: (3×12) + (8×1) = 44 g/mol

mass = mole × molar mass

mass = 0.6 × 44

mass of C₃H₈ = 28 g

Answer: a) mass of C₃H₈ = 28 g

For alternative b we follow the same process starting from step 3:

3)Now we use the proportion of the balanced equation:

5 mol of O₂ ---- 3 mol of CO₂

x mol of O₂ ----- 1.9 mol of CO₂

x = 3.16 mol of O₂

4) Now we transform mole of O₂ into grams using its molar mass.

molar mass of O₂ is: (2×16) = 32 g/mol

mass = mole × molar mass

mass = 3.16 × 32

mass of O₂ = 101 g

Answer: b) mass of O₂ = 101 g

For alternative c we follow the same process starting from step 3:

3) Now we use the proportion of the balanced equation:

4 mol of H₂O ---- 3 mol of CO₂

x mol of H₂O ----- 1.9 mol of CO₂

x = 0.84 mol of H₂O

4) Now we transform mole of H₂O into grams using its molar mass.

molar mass of H₂O is: (2×1) + (1×16) = 18 g/mol

mass = mole × molar mass

mass = 0.84 × 18

mass of H₂O = 15 g

Answer: c) mass of H₂O = 15 g

The pH of a basic solution is 8.13. What is [OH⁻]?

Answers

The [OH⁻] of the solution with pH of 8.13 is 1.35 * 10-6 M

pH is the measurement of the acidity or basicity of a compound by measuring [H⁻] ions in the solution. It ranges from 0 to 14 with acidic range from 0 – 7 and basic range from 7-14.

pOH is the measurement of the acidity or basicity of a compound by measuring [OH⁻] ions in the solution.

Thus, pH + pOH = 14

pOH = 14 - pH = 14 – 8.13 = 5.87

pOH = - log ([OH⁻])

- log ([OH⁻]) = 5.87

Log ([OH⁻]) = - 5.87

[OH⁻] = 10 ^ - 5.87   = 0.00000134896

[OH] = 1.35 * 10⁻⁶ M

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What impact did cargo ship refrigeration systems have on the banana industry? Refrigeration slows the rate at which food is being spoiled, refrigeration does not affect speed at which the ship moves, refrigeration does not cause ripening, refrigeration does not control the sugar content of bananas?

Answers

Answer

Refrigeration slows the rate at which food is being spoiled

Explanation

One of the importance of storing foods at cold temperatures (refrigeration) is to slow the growth of microorganisms, thereby limiting food poisoning while preserving food's nutritional qualities and good taste.

Therefore, the impact the cargo ship refrigeration systems have on the banana industry is:

Refrigeration slows the rate at which food is being spoiled


A silver ring is composed of 1.81×1023 atoms. Calculate the mass of the ring in grams.

Answers

Solving with Moles

Mole equation: [tex]n=\dfrac{m}{MM}[/tex]

n = number of moles (mol)m = mass (g)MM = molar mass (g/mol)

To find mass given number of atoms:

Divide number of atoms by the number of atoms in the chemical formula ⇒ find number of moleculesDivide number of molecules by Avogadro's number ([tex]6.02*10^{23}[/tex]) ⇒ find number of moles (n)Solve for m using moles equation

Solving the Question

We're given:

Ag (silver)Atoms = [tex]1.81*10^{23}[/tex] atomsm = ?

In the chemical formula, which is Ag, there is only 1 atom.

Divide [tex]1.81*10^{23}[/tex] atoms by 1 atom to get the number of molecules in the silver ring:

[tex]1.81*10^{23}\div 1\\=1.81*10^{23}[/tex]

Therefore, there are [tex]1.81*10^{23}[/tex] molecules in the silver ring.

Now, divide [tex]1.81*10^{23}[/tex] molecules by Avogadro's number to find n:

[tex]\dfrac{1.81*10^{23}}{6.02*10^{23}}\\\\=\dfrac{1.81}{6.02}\\\\=\dfrac{1.81}{6.02}\\\\= 0.30066[/tex]

Therefore, the sample has 0.30066 mol.

Finally, solve for the mass using the moles equation:

[tex]n=\dfrac{m}{MM}[/tex]

⇒ Rearrange the equation:

[tex]m=n*MM[/tex]

⇒ MM of Ag = 107.87 g/mol

⇒ Plug in given information:

[tex]m=0.30066* 107.87\\m=32.4[/tex]

Therefore, the mass of the ring is 32.4 g.

Answer

32.4 g

Planets A, B, and C revolve around stars like the sun in orbits like that of Earth. Could planet A possibly support human life?

Answers

The planet, known as Proxima b, may be warm enough for liquid water to exist on its surface and hence be suitable for life because it is located within the star's "habitable zone."

The first and second laws of Kepler were what?

Applying Kepler's laws: The sun is at the centre of elliptical planetary orbits, according to the first law. Second Law: The radius vector from the sun to a planet covers the same area in exactly the same amount of time. Third Law: For every planet, there is a constant relationship between the cube of the elliptical semimajor axis and the square of the period of revolution.

Since the angular momentum is changing at a rate of zero, the angular momentum must be constant, which implies that the rate of change of the swept-out area for the celestial body's orbit must also be constant. This leads to Kepler's Second Law, which states that celestial objects in orbit cover equal regions in similar amounts of time.

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Write a balanced equation for the decomposition reaction that occurred in Experiment 2. Include physical states.

Answers

The answer will be in the picture

A gas occupying 0.6 L at 1.70 atm expands to 0.9 L. What is the new pressure assuming temperature remains constant?

Answers

Answer:

1.13 atm

Explanation:

The new pressure can be found by using the formula

P1V1 = P2V2

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we're finding the new pressure P2 we make P2 the subject

We have

[tex]p_2 = \frac{p1v1}{v2} \\ [/tex]

P1 = 1.7 atm

V1 = 0.6 L

V2 = 0.9 L

We have

[tex]p_2 = \frac{1.7 \times 0.6}{0.9} = 1.13333...\\ [/tex]

We have the final answer as

1.13 atm

Hope this helps you

Can someone draw this for me?
Btw you don't need real gumdrops or toothpicks you just need to draw it.
I'll mark the brainliest.

Answers

Both calcium oxide and carbon dioxide are inorganic compounds where calcium oxide is an ionic compound with a positive end at calcium and negative end at oxygen and carbon dioxide is a covalent compound formed by two double bonds between carbon and each oxygen.

What is calcium oxide?

Calcium oxide is also called soda lime which is an inorganic compound formed by the ionic bonding of oxygen and calcium. Calcium is a metal and it is electron rich with 2 valence electrons.

Oxygen is electronegative gas, having 6 valence electrons and require two electrons to be stable. Thus calcium donates its two electrons to oxygen making calcium a positive ions and oxygen a negative ion.

In carbon dioxide, carbon have a valency of four thus, it forms 4 bonds with two oxygens. Double bonds are formed with each oxygen as shown in the image. These bonds are formed by electron sharing and is called covalent.

Calcium carbonate is formed by passing carbon dioxide through lime (CaO) in which Ca exists as Ca²⁺ ion and carbonate as an anion with three negative charges residing on 2  oxygens in CO₃.

Therefore, the structure of each of these compounds depends on the type of bonds between the atoms and it is depicted in the uploaded image.

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17 was found from determination in a mass spectrometer that an element X has three Isotopes whose mass are & 19.19,20.99 and 21.99 respectively The abundance of these I sotopes are 90.92% 0.25% $8.83% respectively Calculate the relative atomic mass.​

Answers

The relative atomic mass of the given element is 40.372 amu.

What is relative atomic mass?

The relative atomic mass of an element is considered as the sum of the isotopes masses each multiplied by the percentage which is found in nature.

The formula which is used to calculate the relative atomic mass is

Relative atomic mass = sum of all atomic masses of isotopes × fractional abundance

Given,

Mass of isotopes 1 = 19.19 amu

Mass of isotopes 2 = 20.99 amu

Mass of isotopes 3 = 21.99 amu

Fractional abundance of isotope 1 = 0.9092

Fractional abundance of isotope 2 = 0.0025

Fractional abundance of isotope 3 = 0.0883

By substituting all the values, we get

[( 19.19 × 0.9092) + (20.99 × 0.0025) + (21.99 × 0.883)]

= 17.447 + 0.052 + 22.873

= 40.372 amu.

Thus, we concluded that the relative atomic mass of the given element is 40.372 amu.

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the atomic masses of the two stable isotopes of beam

Answers

Answer:

10.81amu

Explanations:

In order to get the average atomic mass of an element, we need the following parameters:

• Natural Abundance (NA),: The percentage of atoms for an element that is a specific isotope.

• Mass (m) ,of each isotope

For the given element (Boron-10 and Boron-11), the natural abundances are 19.78% and 80.22% respectively.

The atomic masses of Boron-10 and Boron-11 are 10.0129amu and 11.0093amu respectively

The formula for calculating the average atomic mass of the element is expressed as:

[tex]AAM=(NA_a\times m_a)+(NA_b\times m_b)[/tex]

Substitute the given parameters into the formula to have:

[tex]A\mathrm{}A\mathrm{}M=(0.1978\times10.0129)+(0.8022\times11.0093)[/tex]

Simplify the resulting expression to have:

[tex]\begin{gathered} A\mathrm{}A\mathrm{}M=1.98055162+8.83166046 \\ A\mathrm{}A\mathrm{}M=10.81221208 \\ A\mathrm{}A\mathrm{}M\approx10.81amu \end{gathered}[/tex]

Therefore the average atomic mass of Boron is 10.81amu to two decimal places.

Write formulas or names as appropriate for each of the following ionic compounds. 1. Magnesium nitride 6. SrI2 2. Lithium oxide 7. Ba3(PO4)2 3. Aluminum sulfite 8. (NH4)2O 4. Copper(II) bicarbonate 9. Fe(ClO)3 5. Sodium nitrate 10. ZnCrO4

Answers

We have the following formulas for the given compounds:

1. Magnesium nitride ---> Mg3N2

2. Lithium oxide ---> Li2O

3. Aluminum sulfite ---> Al2(SO4)3

4. Copper (II) bicarbonate ---> Cu(HCO3)2

5. Sodium nitrate ---> NaNO3

For the given formulas we have the following names:

6. SrI2 ---> Strontium iodide

7. Ba3(PO4)2 --->Barium phosphate

8. (NH4)2O ---> Ammonium oxide

9. Fe(ClO)3 ---> Iron(III) hypochlorite

10. ZnCrO4 ---> zinc chromate

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