If the oh- of a solution is 2.7x10-4m the poh of the solution is

Answer: Look up a video, and they will tell/show you.

Answer:

A

Explanation:

.75 + .25 + 1.25 = 2.25 atm

1 atm is 760 mm hg

2.25 * 760 = 1710 mm HG

Answer:

[tex]\huge\boxed{\sf 2.25\ atm = 1710\ mm\ of\ Hg}[/tex]

Explanation:

Partial pressure of gas A = 0.75 atm

Partial pressure of gas B = 0.25 atm

Partial pressure of gas C = 1.25 atm

Total partial pressure = 0.75 atm + 0.25 atm + 1.25 atm

= 2.25 atm

We know that:

1 atm = 760 mm of Hg

Multiply 2.25 to both sides

2.25 atm  = 760 × 2.25 mm of Hg

2.25 atm = 1710 mm of Hg

[tex]\rule[225]{225}{2}[/tex]

Answer:

heat carbunate

Explanation:

thats true im good in here like and heart byebye

Answer:

the answer is 12

Explanation:

In one formula unit of Al2(SO4)3 A l 2 ( S O 4 ) 3 , there would be twelve atoms of oxygen.

Answer:

a

Explanation:

On drawing the structure, the most Carbon atoms in a row is 5

The remaining two carbon atoms are connected to the 2nd carbon atom

IUPAC name is :

Answer:

Explanation:

You have the equation. Now change the 3.4 g H2 to moles. moles = grams/molar mass

3.4 g/2.016 = 1.686 moles.

Now using the coefficients in the balanced equation, convert moles H2O2 to moles H2O.

1.686 moles H2 x (2 moles H2O/2 moles H2O2) = 1.686 x (2/2) = 1.686 x (1/1) = 1.686 moles H2O.

Now you know that 1 mole of water is composed of 6.022 x 10^23 molecules. So

1.686 moles H2O x (6.022 x 10^23 molecules H2O/1 mole H2O) = ?? molecules.

Answer:

The answer is Polar.

Explanation:

Because polar has many unique properties.

The concentration of a substance can be expressed using molarity, molality or percent.

The term concentration has to do with the amount of substance present in solution. Now let us solve the problems individually;

a) Number of moles =  0.850 grams/80 g/mol = 0.011 moles

molarity = 0.011 moles/345 * 10^-3 L = 0.032 M

b) Number of moles =  13.5g /58 g/mol = 0.23 moles

molality = 0.23 moles/250 * 10^-3 Kg = 0.92 m

c) Number of  moles = 16.5 g/128 g/mol = 0.13 moles

molality = 0.13 moles/54.3  * 10^-3 Kg =2.39 m

d) Total mass present =  12 g +  18 g + 25 g + 500 g = 555 g

mass percent of calcium sulfate =  12 g/555 g * 100/1 = 2.2 %

mass percent of sodium nitrate =  18 g/ 555 g  * 100/1 = 3.2 %

mass percent of potassium chloride = 25 g / 555 g  * 100/1 = 4.5%

mass percent of water =  500 g / 555 g  * 100/1 =90.1%

e) Total mass present = 125 g + 1500g = 1625 g

Mass percent of NaCl =  125 g/1625 g * 100/1 = 7.7%

f) Total volume of solution = 15 L +  28 L  = 43 L

percent by volume of acetone = 15 L/43 L * 100/1 =  34.9%

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Answer:

3) The amount of matter within the system remains the same

5) The cycle has a well-defined starting and stopping point.

Explanation:

A closed system is a system that the boundaries of a system cannot be passed by either matter or energy.

Hence, generalisations can be made based on a closed system and the law of conservation of mass which states that matter can neither be created nor destroyed but can only change form.

The mass of fluorine that were formed is 592 mg

From the question, we are to determine the mass of fluorine formed

First, we will write the balanced chemical equation for the decomposition reaction

The balanced chemical equation for the reaction is

CaF₂ → Ca + F₂

This means

1 mole of calcium fluoride decomposes to give 1 mole of calcium and 1 mole of fluorine

Now, we will determine the number of moles of calcium produced

From the given information,

Mass of calcium produced = 625 mg = 0.625 g

Using the formula,

[tex]Number\ of \ moles =\frac{Mass}{Atomic\ mass}[/tex]

Atomic mass of calcium = 40.078 g/mol

Then,

Number of moles of calcium produced = [tex]\frac{0.625}{40.078}[/tex]

Number of moles of calcium produced = 0.01559 mole

Since

1 mole of calcium fluoride decomposes to give 1 mole of calcium and 1 mole of fluorine

Then,

0.01559 mole of calcium fluoride will decompose to give 0.01559 mole of calcium and 0.01559 mole of fluorine

∴ Number of mole of fluorine formed was 0.01559 mole

Now, for the mass of fluorine formed

Using the formula,

Mass = Number of moles × Molar mass

Molar mass of fluorine = 38 g/mol

Then,

Mass of fluorine formed = 0.01559 × 38

Mass of fluorine formed = 0.59242 g

Mass of fluorine formed = 592.42 mg

Mass of fluorine formed ≅ 592 mg

Hence, the mass of fluorine that were formed is 592 mg.

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The equation that shows how frequency is related to velocity and wavelength is f = v / λ (Option A)

This is defined as the number of oscillations made in one second. It is expressed as

Frequency (f) = 1 / period (T)

f = 1 / T

The velocity, frequency and wavelength of a wave are related according to the following equation

v = λf

Making f the subject, we have

f = v / λ

Thus, we can conclude that the correct answer to the question is:

f = v / λ (Option A)

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So

Moles of Oxygen:-

3 mol of O_2 produces 2 mol SO_2

1 mol of ao_2 produces 2/3=0.6mol So_2

1.396 mol produces 0.8376mol SO_2

Mass of SO_2

The molecular formular shows all the atoms in the compound. The molecular formula of the compound is C15H10O5.

The molecular formula is the formula of the compound that shows the total number of atoms in a compound.

We know that;

ΔT = K m i

ΔT = 26.4°C

K = 40.0°C kg/mol

m = ?

i = 1

26.4°C = 40.0°C kg/mol * m * 1

m = 26.4°C/40.0°C kg/mol  = 0.66 m

We know that

m = mass/molar mass ÷ mass of solvent(in Kg)

Let the molar mass be MM

m = 3.044 g/MM ÷ (18 * 10^-3 Kg)

0.66 m =  3.044 g/0.018MM

0.66 m  * 0.018MM =  3.044 g

MM = 3.044 g/0.66 m  * 0.018

MM = 256 g/mol

Now;

Mass of C =  17.80 * 10^-3/44 * 12 = 0.00485 g

Moles of C = 0.00485 g/12 g/mol = 0.0004 moles

Mass of H = 2.08 * 10^-3/18  * 2 = 0.0002 g

Moles of H =  0.0002 g/1 g/mol = 0.0002 moles

Mass of O = [7.40 * 10^-3 - (0.00485 + 0.0002)] = 0.00235 g

Moles of O = 0.00235 g/16 g/mol = 0.000147 moles

Divide through by the lowest number of moles;

C - 0.0004/0.000147  H - 0.0002/0.000147  O - 0.000147 /0.000147

C - 3   H  -  2  O -  1

The empirical formula is C3H2O

So;

[3(12) + 2(1) + 1(16)]n = 256

[36 + 2 + 16]n = 256

n = 5

Molecular formula = C15H10O5

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Answer:

True

Explanation:

Answer:

true

Explanation:

Answer:

Plant Adaptations is a unique feature a plant has that allows it to live and survive in its own particular habitat (the place that it lives). Desert Plant Adaptations – Plant Adaptation is really a unique have a plant has that enables it to reside and survive in the own particular habitat (the area it lives).

Explanation:

From the equation of te reaction, we know that the mass percent  of NaOH in the mixture is 1.4%.

Neutralization is a reaction that occurs between an acid and a base to yield salt and water only.

In tis case, the reaction of the NaOH and HCl occurs as follows; NaOH + HCl ----> NaCl + H2O

Number of moles of HCl reacted = 15/1000 * 0.5 moldm-³ = 0.0075 moles

Since the reaction is 1:1, 0.0075 moles of NaOH reacted.

Mass of NaOH =  0.0075 moles of NaOH * 40 g/mol = 0.3 g

Percent of NaOH = 0.3 g/21.10g * 100/1 = 1.4%

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Answer:

B

Explanation:

A mixture in which the composition is not uniform throughout.

Same thing letter B says.

The freezing points of each of the solutions are as follows;

0.100 m K2S -   - 0.558oC

21.5 g of CuCl2 in 4.50 * 102 g water -   -2oC

5.5% NaNO3 by mass (in water)​ -     - 2.6oC

The freezing point is the point at which liquid changes to solid. Let us now look at the freezing point of each solution.

a)

Since;

ΔT = K m i

K = 1.86 oC m-1

m = 0.100 m

i = 3

ΔT =   1.86 oC m-1 * 0.100 m * 3 = 0.558oC

Freezing point = 0oC - 0.558oC =  - 0.558oC

b) Number of moles of CuCl2 = 21.5 g/134.45 g/mol = 0.16 moles

molality = 0.16 moles/0.45 Kg = 0.36 m

ΔT = K m i

ΔT =  1.86 oC m-1 * 0.36 m * 3 = 2oC

Freezing point = 0oC - 2 = -2oC

c) Number of moles of NaNO3 = 5.5g/85 g/mol = 0.065 moles

molality of the solution =  0.065 moles/0.0945 Kg = 0.69 m

ΔT =  1.86 oC m-1 * 0.69 m * 2 = 2.6oC

Freezing point = 0oC - 2.6oC = - 2.6oC

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The ratio of moles of AI(NO3)3 to moles Na2CO3 in the reaction is 2:3

In the reaction 2 moles of AI(NO3)3 reacted with 3 moles Na2CO3 so the ratio of the moles is 2 is to 3 represented as 2:3

hope it helps

False, there are 9 named classes of hazardous materials.

Hazardous materials are substances or chemicals that pose a health hazard, a physical hazard, or harm to the environment.

There are 9 hazardous substances symbols you need to know: flammable, oxidising, explosives, gas under pressure, toxic, serious health hazard, health hazard, corrosive and environmental hazard.

Hence, the statement is false.

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Based on the charge on the aluminium ion, 0.9 g of aluminium are deposited by 0.1 F of electricity.

Electrolysis is the decomposition of a substance known as an electrolyte when electric current is passed through it.

The mass and hence moles an electrolyte deposited when current is passed through it depends on the charge on the ion.

Aluminium ion has a charge of +3 and requires 3F of electricity to deposit 1 mole or 27 g of aluminium

0.1 F will discharge = 0.1/3 × 27 g of aluminium

mass of aluminium deposited = 0.9 g of aluminium.

Therefore, 0.9 g of aluminium are deposited by 0.1 F of electricity.

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Answer:

gaseous state.

Explanation:

that's the answer

Answer:

Depending on the actual amount of water, it's going to overfill.

Explanation:

If your watershed is already one that has a lot of water and is always filled, it's likely that your wetlands and rivers will start to overfill, causing a form of flooding.

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I hope this helps!

-No one

Answer:

About 0.164 g of glucose.

Explanation:

We can determine the mass of glucose produced given the volume of oxygen gas produced with stoichiometry.

Recall that at STP, a mole of any gas occupies a volume of 22.4 L.

From the reaction, six moles of oxygen gas is produced for every one mole of glucose.

Lastly, the molecular weight of glucose is 180.18 g/mol.

Therefore:
[tex]\displaystyle \begin{aligned} 122\text{ mL O$_2$} & \cdot \frac{1\text{ L O$_2$}}{1000\text{ mL O$_2$}}\cdot \frac{1\text{ mol O$_2$}}{22.4\text{ L O$_2$}} \cdot \frac{1\text{ mol C$_6$H$_{12}$O$_6$}}{6\text{ mol O$_2$}} \cdot \frac{180.18\text{ g C$_6$H$_{12}$O$_6$}}{1\text{ mol C$_6$H$_{12}$O$_6$}} \\ \\ & = 0.164\text{ g C$_6$H$_{12}$O$_6$}} \end{aligned}[/tex]

Therefore, about 0.164 g of glucose is produced.

The specific heat of a mystery substance that requires 79 joules of energy to heat 4.52 grams of substance from 23 degrees Celsius to 54 degrees Celsius is 0.56 J/g°C.

The specific heat capacity of a substance can be calculated using the following formula:

Q = mc∆T

Where;

c = Q/m∆T

c = 79 ÷ 4.52 × (54 - 23)

c = 79 ÷ 140.12

c = 0.56 J/g°C

Therefore, the specific heat of a mystery substance that requires 79 joules of energy to heat 4.52 grams of substance from 23 degrees Celsius to 54 degrees Celsius is 0.56 J/g°C.

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Principle: Orbital size & main energy level

Angular: Orbital shape

Magnetic: Orbital orientation

Orbital size is Principle; main energy level is Principle; shape is Angular; orientation is Magnetic

Principle:

imagine a bullseye. each ring shows how far its from the nucleus

Angular:

0 looks like circle

1 looks like a bowtie

2 looks like a flower or 4 leaf clover

3 looks like multiple lobes

Principle Quantum Number: It describes the size of the orbital and the main energy level. It is represented by n. Where, n = 1,2,3,4....stands for K, L, M, N shell and so on..

Angular momentum Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as . The value of this quantum number ranges from . When l = 2, the value of  will be -2, -1, 0, +1, +2.

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The specific heat of an unknown substance is 0.96 J/(g·ºC).

The specific heat capacity of a substance can be described as the quantity of heat needed to raise the temperature in one unit of substance by 1 degree Celcius.

Whenever the heat is lost or absorbed there is a change in the temperature of the substance:

Q = mCΔT

Given, the amount of energy released, Q = 12 cal

The initial temperature of the substance = 25.0ºC

The final temperature of the substance = 20.0ºC

The change in the temperature of a substance, ΔT = 25 -20 = 5°C

The mass of the substance, m = 2.50 g

We have to find the specific heat capacity of an unknown substance.

12 cal = 2.50 g × C × 5°C

C = 0.96 J/(g·ºC)

Therefore, the specific heat capacity of an unknown substance is equal to 0.96 J/(g·ºC).

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The amount of heat needed to raise a substance's temperature by one degree Celsius per gram is known as specific heat. The specific heat in J/(gºC) of an unknown substance is 0.96 J/gºC.

The amount of heat needed to increase the temperature of the entire substance by one degree is the heat capacity of a substance. Specific heat capacity is what is used when the substance's mass is equal to one.

The hydrogen bonds in water, which account for its high specific heat.

The expression used to calculate the specific heat capacity is:

Q = mcΔT

Here Q = 12 Cal

m = 2.50 g

T₁ = 25.0ºC

T₂ = 20.0ºC

ΔT = 5°C

12 cal = 2.50 g × c × 5°C

c = 0.96 J/gºC

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Answer:

the answer has been given below have a good day

Explanation:

mark as brainiest

The species are present at 10-6 mole/L or greater when dissolved in water is when in solutions, weak acids and bases partially ionize.

Ionization is defined as anything that causes electrically neutral atoms or molecules to gain or lose electrons in order to become electrically charged atoms or molecules (ions).  One of the main mechanisms by which radiation, including charged particles and X-rays, transmits their energy to matter is ionization. An anion is created when an atom or molecule picks up an electron; a cation is created when they lose an electron.

Weak Acids and Weak Bases Ionizing. Numerous acids and bases are weak, meaning they do not completely ionize in aqueous solution. The nonionized acid, hydronium ion, and conjugate base of a weak acid are all mixed together to form a solution in water, with the nonionized acid being present in the highest concentration.

Thus, the species are present at 10-6 mole/L or greater when dissolved in water is when in solutions, weak acids and bases partially ionize.

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