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2 years ago

Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.

Chemistry

1 answer:

Maksim231197 [3]2 years ago

7 0

The theoretical yield of $SnS_2$ will be 4.20 grams while the percent yield will be 7.93%

<h3>How is yield calculated?</h3>

From the equation of the reaction, the mole ratio of $SnBr_4$ to $Na_2S$ is 1:2.

Mole of 48.1 mL, 0.478 M $SnBr_4$ = 0.478 x 48.1/100 = 0.023 mols

Mole of 48.8 mL, 0.160 M $Na_2S$ = 0.160 x 48.8/1000 = 0.0078 moles

$SnBr_4$ $Na_2S$ is the limiting reactant.

Mole ratio of $SnBr_4$ and $SnS_2$ = 1:1

Equivalent mole of $SnS_2$ = 0.023 moles

Mass of 0.023 noles $SnS_2$ = 0.023 x 182.81 = 4.20 grams

With 0.0333 g of $SnS_2$ recovered, percent yield = 0.333/4.2 x 100 = 7.93%

More on yields of reactions can be found here: brainly.com/question/17042787

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Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.

SnBr4(aq)+2Na2S(aq)⟶4NaBr(aq)+SnS2(s)

Suppose a student adds 48.1 mL of a 0.478 M solution of SnBr4 to 48.8 mL of a 0.160 M solution of Na2S.

1) Calculate the theoretical yield of SnS2. ;

2) The student recovers 0.333 g of SnS2. Calculate the percent yield of SnS2 that the student obtained.

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I looked this up but hope it helps!

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3 years ago

What charge will an ion of phosphorus form?

Fed [463]

Answer:

A charge of -3

Explanation:

If you look on a periodic table you will find that phosphorus is 3 elements away from the nearest noble gas argon. This means that phosphorus would prefer a charge of -3!

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3 years ago

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How many atoms are in 25.00 g of B?

klio [65]

Answer:

There are 1.393 x 10²⁴ atoms in 25.00 g of B.

Explanation:

Hey there!

We are given a value, in grams, that we need to convert to a number of atoms.

We can convert grams to atoms by using Avogadro's Number ( $N_A$ ). This number is equivalent to $6.022 \times 10^{23}$ .

This number can be used to convert any values to:

atoms
molecules
formula units
moles

In order to do this problem, we will need to use dimensional analysis (DA). This process allows us to convert from grams to atoms.

We need to set up our ratios in order to work this out. We can use a periodic table to help us through this next part of the problem.

<u>1. Locating the number of moles of B in the sample</u>

We first need to find the amount of moles of boron (B) there are in the sample.

Checking a periodic table, the atomic mass in atomic mass units (amu) is 10.81 amu.

Atomic mass units can easily be converted to grams and these units can be used interchangeably.

Therefore, for each atom of boron, it weighs 10.81 grams to us. This is equivalent to the mass of one mole of boron.

To find the number of moles, we have two possible ratios we can use:

$\displaystyle \frac{1 \ mole \ B}{10.81 \ grams \ B}$

$\displaystyle \frac{10.81 \ grams \ B}{1 \ mole \ B}$

These ratios mean the same thing, but we need to convert our final unit to moles.

We are given a sample in grams, and when dividing our units, we need to keep moles.

Since the first portion of our expression is in grams, we need to have grams in the bottom of our expression.

$\displaystyle 25.00 \ \text{grams B} \ \times \frac{1 \text{mole B}}{10.81 \ \text{grams B}}$

We can now simplify the expression. Our <u>grams B</u> unit will cancel out, so we are therefore left with <u>moles B</u> remaining.

<u>2. Locating the number of atoms in the sample</u>

Now with our equation, we can convert our number of moles that would be solved if we stopped with the above. However, we need to convert to atoms.

We use Avogadro's number and create a ratio with that of moles.

$\displaystyle \frac{6.022 \times 10^{23}\text{atoms}}{1 \text{mole B}}$

$\displaystyle \frac{1 \text{mole B}}{6.022 \times 10^{23} \text{atoms}}$

We need to cancel out our moles and end with atoms, so we must have moles in the denominator. Therefore, we use the first ratio.

Using our previous expression, we multiply by this new ratio and solve the expression.

$\displaystyle 25.00 \ \text{grams B} \ \times \frac{1 \text{mole B}}{10.81 \ \text{grams B}} \ \times \frac{6.022 \times 10^{23}\text{atoms}}{1 \text{mole B}}$

This expression can now be operated. You will need a calculator to perform this calculation.

<u>Our numerator is:</u>

$[(25.00 \times 1 \times (6.022 \times 10^{23})]$

Plugging this into a calculator, we get:

$1.5055 \times 10^{25}$

<u>Our denominator is:</u>

$(1 \times 10.81 \times 1)$

This simplifies to:

$10.81$

<u>Dividing our numerator and denominator:</u>

<u /> $\displaystyle \frac{1.5055 \times 10^{25}}{10.81}$ <u />

Plugging this into a calculator, we get:

$1.392691952 \times 10^{24}$

<u>3. Simplifying with significant figures</u>

Now, we need to take into account that we have significant figures. We are given this original value:

$25.00$

This value has four significant figures, which means we need to round our value we received above to four significant figures.

$\approx 1.393$

Our units are added as well as our scientific notation:

$1.393 \times 10^{24} \ \text{atoms of B}$

Therefore, our final answer is choice A.

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3 years ago

A certain gas is found in the exhaust of automobiles and power plants. Consider a 1.53 L sample of a gas at a pressure of 5.6 x

Olegator [25]

The new volume of the gas that has an initial pressure of 5.6 x 10⁵ Pa is 5.7L. Details about volume of the gas can be found below.

<h3>How to calculate volume?</h3>

The volume of a gas can be calculated using the Boyle's law equation as follows:

P1V1 = P2V2

Where;

P1 = initial pressure
P2 = final pressure
V1 = initial volume
V2 = final volume

1.53 × 5.6 × 10⁵ = 1.5 × 10⁵ × V2

8.568 × 10⁵ = 1.5 × 10⁵V2

V2 = 8.57 × 10⁵ ÷ 1.5 × 10⁵

V2 = 5.7L

Therefore, the new volume of the gas that has an initial pressure of 5.6 x 10⁵ Pa is 5.7L.

Learn more about volume at: brainly.com/question/12357202

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2 years ago

Which of the following statements is true?

hammer [34]

Lord dragon you are wrong.

The answer is A: Incandescent light bulbs are almost 100 percent efficient. Because b-d do not make any sense.

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3 years ago

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