Answer:
Both are caused by heating. Both are Chemical changes.
Explanation:
Water evaporates from the Sun's heat and is transformed into a gas. Sediment is the particles of an object. When heat hits it, it starts breaking up, or in dry cases, heating and cracking.
2. A 0.2719 g sample containing CaCO3 reacted with 20.00 mL of 0.2254 M HCl. Given that HCI was excess. The excess HCl required exactly 20.00 mL of 0.1041 M NaOH to reach the end-point using phenolphthalein indicator. Determine percentage purity of CaCO3 in the sample. The reraction involved is CaCO3(s) + 2HCl(aq) → CaCl₂(aq) + 2H₂O(l) The titration reaction is HC(aq) + NaOH(aq) NaCl(aq) + H₂O(1)
The calculated percentage purity of the original sample of calcium carbonate is 44%.
What is excess titration?In the back titration of the excess acid, we neutralize the excess titrand left in a system.
We have the reaction equation;
CaCO3 + 2HCl ----> CaCl2 + H2O + CO2
To get the moles of HCl = 0.2254 M * 20/1000 L = 0.0045 moles
The reaction involving the excess acid occurs thus;
HCl + NaOH ----> NaCl + H2O
Number of moles of NaOH = 0.1041 M * 20/1000 = 0.0021 moles
We have a 1:1 hence 0.0021 moles of HCl reacted also
Number of moles of HCl that reacted with CaCO3 initially = 0.0045 moles - 0.0021 moles = 0.0024 moles
Given the stoichiometry of the reaction, 0.0012 moles of CaCO3 reacted.
Hence;
Mass of pure CaCO3 in the sample= 0.0012 moles * 100 g/mol = 0.12g
Percent purity of the sample = 0.12g/0.2719 g * 100/1
= 44%
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Which of the following is the correct definition of nuclear energy? A. energy of an object due to the random motion of its atoms and molecules B. energy an object has because of its motion or position C. energy produced from the splitting of atoms D. energy resulting from the flow of charged particles, such as electrons or ions
Answer:
c. energy from splitting energy
Explanation:
Determine the hybridization about O in CH3oh
Answer:
c
Explanation:
lol
How much energy (in J) is required to raise the temperature of 19.2 g of water by 5.00 oC? (Specific heat capacity of liquid water: 4.184 J/g⋅°C)
2. A 0.2719 g sample containing CaCO3 reacted with 20.00 mL of 0.2254 M HCl. Given that HCI was excess. The excess HCl required exactly 20.00 mL of 0.1041 M NaOH to reach the end-point using phenolphthalein indicator. Determine percentage purity of CaCO3 in the sample. The reraction involved is CaCO3(s) + 2HCl(aq) → CaCl₂(aq) + 2H₂O(l) The titration reaction is HC(aq) + NaOH(aq) NaCl(aq) + H₂O(1)
The percentage purity of the calcium carbonate is 44%. The percentage purity gives the amount of pure CaCO3 in the sample.
What is excess titration?In excess titration or back titration, we neutralize the excess titrand left in a system.
We have the reaction;
CaCO3 + 2HCl ----> CaCl2 + H2O + CO2
Number of moles of HCl = 0.2254 M * 20/1000 L = 0.0045 moles
The reaction of the excess acid is according to the reaction;
HCl + NaOH ----> NaCl + H2O
Number of moles of NaOH = 0.1041 M * 20/1000 = 0.0021 moles
Since the reaction is 1:1, 0.0021 moles of HCl reacted also
Number of moles of HCl that reacted with CaCO3 = 0.0045 moles - 0.0021 moles = 0.0024 moles
If 2 moles of HCl reacts with 1 mole of CaCO3
0.0024 moles of HCl reacts with 0.0024 moles * 1/2
= 0.0012 moles
Mass of pure CaCO3 present = 0.0012 moles * 100 g/mol = 0.12g
Percent purity of the sample = 0.12g/0.2719 g * 100/1
= 44%
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Brainliest
Which choice is an example of the use of sound energy in everyday life?
batteries powering toys
fireplaces keeping people warm
lights helping people see
car horns preventing accidents
Answer:
car hornsss
Explanation:
they loud asl
What is the volume of hydrogen chloride under normal conditions, if its mass is 73 g?
Explanation:
36.5 g of HCl occupies 2.24×10-2m3 . ∴ 73 g of HCI at STP will occupy 2.24×10-236.5×732=4.48×10-2m3.
Write a balanced equation for the precipitation reaction that was used to separate Ba2+ from Ca2+ and Mg2+.
Answer:
[tex]{\rm Ba^{2+}} \, (aq) + {\rm {SO_4}^{2-}}\, (aq) \to {\rm BaSO_{4}}\, (s)[/tex].
Explanation:
Make use of the fact that calcium sulfate [tex]{\rm CaSO_{4}}[/tex] and magnesium sulfate [tex]{\rm MgSO_{4}}[/tex] are much more soluble in water than barium sulfate [tex]{\rm BaSO_{4}}[/tex].
When sulfate ions [tex]{\rm {SO_{4}}^{2-}}[/tex] are added to dilute solutions containing [tex]{\rm Ba^{2+}}\, (aq)[/tex], [tex]{\rm Ca^{2+}}\, (aq)[/tex], and [tex]{\rm Mg^{2+}}\, (aq)[/tex], precipitation would be visible only in the solution with [tex]{\rm Ba^{2+}}\, (aq)\![/tex]. Barium sulfate would be the precipitate.
[tex]{\rm Ba^{2+}} \, (aq) + {\rm {SO_4}^{2-}}\, (aq) \to {\rm BaSO_{4}}\, (s)[/tex].
This ionic equation is balanced as it conserves both the atoms and the charges on the ions.
How many kilograms are present in 6.78x1025molecules of Al2(SO4)3? (hint 1kg=103g)
Atomic masses:
Al = 27
S = 32
O = 16
A) 0.891kg
B) 38.5kg
C) Cannot be determined
D) 8.44kg
mol = 6.78 x 10²⁵ : 6.02 x 10²³ = 112.62
MM Al₂(SO₄)₃ = 2.27 + 3.32+12.16 =342 g/mol
mass = 112.62 x 342 = 38516.04 g = 38.5 kg
The total mass of 6.78*[tex]10^{25}[/tex] molecules of [tex]Al_{2}(SO_{4})^3[/tex] is (b) 38.5 Kg.
To calculate the mass, we need to first understand the values given in the question:1 mole of [tex]Al_{2}(SO_{4})^3[/tex] contains 6.02*[tex]10^{23}[/tex] molecules6.78*[tex]10^{25}[/tex]/6.02*[tex]10^{23}[/tex] moles are in the given amount of [tex]Al_{2}(SO_{4})^3\\[/tex]112.62 moles of [tex]Al_{2}(SO_{4})^3[/tex] is availableWeight of 1 mole [tex]Al_{2}(SO_{4})^3[/tex] is 342 gramsHence, weight of 112.62 moles of [tex]Al_{2}(SO_{4})^3[/tex] = 112.62*342 gramsThe total mass is equal to 38516.04 grams, which is equal to 38.5 Kg.Hence, the total weight of 6.78*[tex]10^{25}[/tex] molecules of [tex]Al_{2}(SO_{4})^3[/tex] is (b) 38.5 Kg.
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Mass of 1.5 mole electron is
Mass of 1.5 mole of electron is 0.8229 mg
mass of electron = 9.11 * 10^-31 kg
mass of 1 mole of any substance ( Avagadro number ) = 6.022 * 10^23 /mol
1.5 moles of electron = 1.5 * 6.022 * 10^23 = 9.033 * 10^23 /mol
mass of 1.5 mole of electron = 9.033 * 10^23 * 9.11 * 10^-31
mass of 1.5 mole of electron = 8.229 * 10^-7 kg
mass of 1.5 mole of electron = 8.229 * 10^-4 g
mass of 1.5 mole of electron = 0.8229 mg
Hope it helps.....
where does the beta particle come from
How many grams of air are required to complete the combustion of 186 g of phosphorus (M= 31 g/mol) to diphosphorus pentoxide, assuming the air to be 23% oxygen by mass?
4P(s) + 5O2(g) — 2P2Os(s)
Answer:
1043.5 g of air to the nearest tenth.
Explanation:
4P + 5O2 ---> 2P2O5
4*31 g P reacts with 5*32 g oxygen
124 g .. .. . .... . 160g oxygen
186 g .. .. .. .. .. (160*186)/124 g oxygen
- so the mass of air required = [(160*186)/124] / 0.23
= 240 / 0.23
= 1043.5 g of air.
Predict: How do you think you could use the Gizmo to make the color gray? _____________
_________________________________________________________________________
Observe: Change the Red, Green, and Blue values to 90. What color does this produce?
_________________________________________________________________________
The simplest way to make the color gray is by mixing white and black colors, but it can also be created by mixing colors such as red and green.
What is the grey color?This is considered a neutral and cool color that can be naturally found in elements such as rocks.
How can you make this color by mixing other colors?Neutral gray: The eeasiest way to create this color is by mixing black and white. This would give you a very neutral gray.Other shades: It is also possible to make this color by mixing colors that are opposite in the colors scale such as:Yellow and purples.Red and green.This would not be a neutral gray as it will resemble a little bit the original colors.
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A gas occupies a volume of 50.0 mL at 27°C. At what temperature, in °C, would the pressure be
101.3 kPa if the volume remains constant?
Answer:
i believe 223
Explanation:
T2 + 500k = 223 C*
Nitric acid (NO) reacts with oxygen gas to form nitrogen dioxide (NO2), a dark brown gas:
2NO(g) + O2(g) → 2NO2(g)
In one experiment, 0.886 mole of NO is mixed with 0.503 mol of O2.
Calculate which of the two reactants is the limiting reagent. (5 points)
Calculate also the number of moles of NO2 produced. (5 points)
What reactant is left over and how much of it is left over? (5 points)
Answer:
Lim reag = NO .07 mole of O2 left over
Explanation:
From the equation, you need twice as many moles of NO as O2
2 x moles O2 = 2 * .503 moles = 1.06 moles of NO needed....you do not have enough so this is the limiting reagent O2 will be left over
1/2 * .866 = .433 moles of O2 used .503 - .433 = .07 mole O2 leftover
. Define the term Flame?
Answer:
A hot glowing body of ignited gas that is generated by something on fire.
25.88 grams of tin (ll) phosphate reacts with 31.73 grams of zinc.
a write a balanced chemical equation
b. what is the limiting reactant?
c. What is the excess reactant and how many moles of it are left over?
d. How many grams of of the products are produced? (2 answers)
Explanation:
Zn + Sn3(PO4)4 = Zn3(PO4)2 + Sn - Balanced Chemical Equation
4.3 is limiting
is the answer
if you like my answer please like comment and mark me as brilliant
What effect would you expect this to have on CO2 levels in the atmosphere?
expect what to have?!?!?!?
How are Jupiter, Saturn, and Uranus similiar? How are Jupiter, Saturn, and Uranus similiar?
Question 5 options:
a They are all large, gaseous planetary bodies.
b They are all planets that are the exact same size.
c They are all the same distance from the sun.
d They all have the exact same surface temperature.
Answer:
a
Explanation:
because they all are gasses
dk how to do this help please
Explanation:
hello
the answer is b
O or oxygen has 16 molar mass and C or carbon has 12
so CO2 or carbon dioxide=(2×O)+C=(2×16)+12=46
Which object will have the greatest gravitational pull?
Question 5 options:
Venus
Sun
Earth
Moon
Answer:
the sun i think because the sun has the greatest gravitational pull
A gas made up of homonuclear diatomic molecules escapes through a pinhole 0.378 times as fast as He gas. Write the chemical formula of the gas.
For a gas made up of homonuclear diatomic molecules escapes through a pinhole 0.378 times as fast as He gas, the chemical formula of the gas is mathematically given as
N2(nitrogen gas)
What is the chemical formula of the gas.?Generally, the equation for the rate to molar relationship is mathematically given as
\frac{r1}{r2}=\frac{mm2}{mm1}
Thereofore
\frac{r}{r2}=0.378
Hence
0.378=\sqrt{\frac{4.00268}{Mx2}}
Mx2=28.013g/mol
In conclusion,The name of gas with a molar mass of 28.013g/mol is N2
Therefore, the chemical formula of the gas. is N2
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Consider the following equilibrium:
N2(g) + 3H2(g) ⇌ 2NH3(g) + 92 kJ
The forward reaction is
Select one:
a.
exothermic and entropy is increasing.
b.
exothermic and entropy is decreasing.
c.
endothermic and entropy is constant.
d.
endothermic and entropy is increasing.
e.
endothermic and entropy is decreasing.
Answer:
b.exothermic and entropy is decreasing
If a car jack's output force is 1000 N and the mechanical advantage is 6.8, what is the input force?
An electromagnet _____________
A. is stronger than a permanent magnet
B. points north
C. can be turned on and off
D. makes noise
I need the answer fast PLS
Answer:
C. can be turned on and off
Explanation:
An electromagnet becomes magnetic when it is part of a complete electric circuit. A switch is a part of an electric circuit that completes or break the circuit turning the circuit on and off.
(*) Sorry for my late answer but I hope this helps others that are looking for this.
I got 100% ;)
How can Climate Change impact the formation of Hurricanes?
I will give brainlist
Order the atoms from largest to smallest: Cs , Si , Sn .
Answer:
Si, Sn, Cs
Explanation:
Atomic no. Atomic radius
Silicon Si_______14___________0.117
Tin Sn______50___________0.140
cesium Cs______55___________0.262
Silicon is the smallest among these
The elements A and B combine to produce two differenr compounds. A3B and AB2. if 0.18 mol of A3B has a mass of 17.7g and 0.18 mol of Ab2 has a mass of 11.3g calculate the molar masses of both elemts a and b.
Taking into account the definition of a system of linear equations and molar mass, the molar masses of elements A and B are 26.778 [tex]\frac{g}{mole}[/tex] and 17.998[tex]\frac{g}{mole}[/tex] respectively.
System of linear equationsA system of linear equations is a set of two or more equations of the first degree, in which two or more unknowns are related.
Solving a system of equations consists of finding the value of each unknown so that all the equations of the system are satisfied. That is to say, the values of the unknowns must be sought, with which when replacing, they must give the solution proposed in both equations.
Definition of molar massThe molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.
The molar mass of a compound (also called Mass or Molecular Weight) is the sum of the molar mass of the elements that form it (whose value is found in the periodic table) multiplied by the number of times they appear in the compound.
Molar masses of elements A and BBeing the molar masses of the compounds:
A₃B: [tex]\frac{17.7 g}{0.18 moles} =[/tex] 98.33 [tex]\frac{g}{mole}[/tex]AB₂: [tex]\frac{11.3 g}{0.18 moles} =[/tex] 62.77 [tex]\frac{g}{mole}[/tex]and considering the definition of molar mass of a compound, the system of equations to be solved is
[tex]\left \{ {{3mA + mB=98.33\frac{g}{mole} } \atop {mA + 2mB=62.77\frac{g}{mole} }} \right.[/tex]
where mA and mB are the molar masses of elements A and B respectively.
There are several methods to solve a system of equations, it is decided to solve it using the substitution method, which consists of clearing one of the two variables in one of the equations of the system and substituting its value in the other equation.
Isolating the variable mB from the first equation:
mB= 98.33 [tex]\frac{g}{mole}[/tex] - 3mA
and substituting in the second equation, you get:
mA + 2× (98.33 [tex]\frac{g}{mole}[/tex] - 3mA)=62.77[tex]\frac{g}{mole}[/tex]
Solving:
mA + 2× 98.33 [tex]\frac{g}{mole}[/tex] - 2×3mA=62.77[tex]\frac{g}{mole}[/tex]
mA + 196.66 [tex]\frac{g}{mole}[/tex] - 6mA=62.77[tex]\frac{g}{mole}[/tex]
mA - 6mA=62.77[tex]\frac{g}{mole}[/tex] - 196.66 [tex]\frac{g}{mole}[/tex]
(-5)mA= - 133.89 [tex]\frac{g}{mole}[/tex]
mA= (- 133.89 [tex]\frac{g}{mole}[/tex] )÷ (-5)
mA= 26.778 [tex]\frac{g}{mole}[/tex]
Remember that mB= 98.33 [tex]\frac{g}{mole}[/tex] - 3mA, replacing the value of mA:
mB= 98.33 [tex]\frac{g}{mole}[/tex] - 3×(26.778 [tex]\frac{g}{mole}[/tex])
mB= 80.334 [tex]\frac{g}{mole}[/tex]
mB= 17.998 [tex]\frac{g}{mole}[/tex]
Finally, the molar masses of elements A and B are 26.778 [tex]\frac{g}{mole}[/tex] and 17.998[tex]\frac{g}{mole}[/tex] respectively.
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A reversible exothermic synthesis (combination) reaction produces one mole of gaseous product from two moles of gaseous
reactant. Which set of condtions will shift the reaction farthest toward the products?
A)
low temperature, low pressure
B)
low temperature, high pressure
9
high temperature, low pressure
D)
high temperature, high pressure
A 2.0g piece of Mg reacts with HNO3. Which conditions would produce the GREATEST reaction rate?
Factors that will increase product formation are;
low temperature and high pressure increasing the surface area and concentrationWhat factors affect chemical equilibrium?The position of chemical equilibrium is affected by the following factors:
temperature concentration pressure for gasesWhen equilibrium is altered by an external factor, the reaction shifts to annul the effect of that change.
For the reversible exothermic synthesis, low temperature and high pressure will shift the reaction farthest toward the products.
For the reaction of the 2.0g piece of Mg with HNO3. increasing the surface area of magnesium metal as well as increasing the concentration of HNO3 would produce the GREATEST reaction rate.
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I need help to balance a Balance the redox reaction under basic conditions Fe+2 +MnO4-1 ---> Fe+3 + Mn+2
Answer:
[tex]5{Fe}^{ + 2} +1{MnO_4}^{ - 1} + 8 {H}^{ + } \rightarrow 5{Fe}^{ + 3} + 1{Mn}^{ + 2} + 4H_2O[/tex]
Explanation:
To balance any redox reaction the primary factor is to identify the species undergoing oxidation and reduction.
Below are the steps to identify species undergoing oxidation and reduction:
Calculate the oxidation number of each atom individually of reactants as well as product.find out the change in oxidation number from reactants to products.if the oxidation number of products are greater than reactants then the species undergone oxidation.if the oxidation number of products are lesser than reactants then the species undergone reduction.let's calculate oxidation number,
given reaction
[tex]{Fe}^{ + 2} +{MnO_4}^{ - 1} \rightarrow {Fe}^{ + 3} + {Mn}^{ + 2}[/tex]
Of reactant,
[tex]Fe = +2, {MnO_4}^{ - 1} \\ Let \: oxidation \: of \: Mn \: is \: x \: and \: Oxygen -2 \\ {MnO_4}^{ - 1} \rightarrow \: x + 4 \times ( - 2) = - 1 \\ {MnO_4}^{ - 1} \rightarrow \: x + - 8 = - 1 \\ {MnO_4}^{ - 1} \rightarrow \: x = + 8 - 1 \\ {MnO_4}^{ - 1} \rightarrow \: x = + 7 \\ \fbox{ Mn \: +7, O \: -2}[/tex]
Of product,
[tex]\fbox{Fe \: +3, \: Mn \: +2}[/tex]
Now calculate the change in oxidation number
[tex]Fe = +3-(+2) = +1 \\ Mn = +2-(+7) = -5[/tex]
now balance the increase in oxidation number and decrease in oxidation number,
thus we require 5 Fe for 1 MnO4-,
Now putting these numbers as their coefficient,
[tex]5{Fe}^{ + 2} +1{MnO_4}^{ - 1} \rightarrow 5{Fe}^{ + 3} + 1{Mn}^{ + 2}[/tex]
now balance oxygen atom on product side by adding water,
since there are 4 oxygen atoms in reactants 4 water molecules will be there on product side.
[tex]5{Fe}^{ + 2} +1{MnO_4}^{ - 1} \rightarrow 5{Fe}^{ + 3} + 1{Mn}^{ + 2} + 4H_2O[/tex]
now balance Hydrogen atom on reactant side by adding H+ ion, since there are 4 water molecules on product 8 H+ ion will be in reactants!
[tex]5{Fe}^{ + 2} +1{MnO_4}^{ - 1} + 8 {H}^{ + } \rightarrow 5{Fe}^{ + 3} + 1{Mn}^{ + 2} + 4H_2O[/tex]
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Confirming the balanced reaction, by checking number of atoms and charge on reactant vs Product side.
Total number of Fe on reactant and product = 5
Total number of Mn on reactant and product= 1
Total number of Oxygen on reactant and product = 4
Total number of hydrogen on reactant and product= 8
Total charge on reactant side= 5×(+2)-1+8=17
Total charge on product side = 5×(+3)+1×(+2)=17
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
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