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2 years ago

I NEED HELP ASAP, WILL MARK BRAINLEST!

Chemistry

1 answer:

Andre45 [30]2 years ago

6 0

Answer:

1. 90%

2. 217.4 g O₂

3. 95.0%

4. Trial 2 ratios

Explanation:

Original: SiCl₄ + O₂ → SiO₂ + Cl₂

Balanced: SiCl₄ + O₂ → SiO₂ + 2Cl₂

Trial SiCl₄ O₂ SiO₂

1 120 g 240 g 38.2 g

2 75 g 50 g 25.2 g

Percentage yield for trial 1

We need to get actual yield (38.2 g) and theoretical yield, in grams.

Mass to moles:

molar mass SiCl₄: 28.09 + 4(35.45) = 169.9 g/mol

120 g SiCl₄ x 1 mol/169.9 g = .706 mol SiCl₄

Moles to moles:

For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

.706 mol SiCl₄ = .706 mol SiO₂

Moles to mass:

molar mass SiO₂: 28.09 + 2(16.00) = 60.09 g/mol

.706 mol SiO₂ x 60.09g/mol = 42.44 g SiO₂

Theoretical yield:

actual/theoretical x 100

38.2 / 42.44 = .900 = 90.0% yield

Leftover reactant for trial 1

We know oxygen is the excess reactant.

Mass to moles:

molar mass O₂ = 32.00 g/mol

240 g O₂ x 1 mol/32.00 g = 7.5 mol O₂

We used .706 mol SiO₂, so we also used .706 mol O₂.

7.5 - .706 = 6.8 moles left over

Moles to mass:

6.8 mol O₂ x 32.00g/mol = 217.4 g O₂

Percentage yield for trial 2

Mass to moles:

molar mass SiCl₄: 169.9 g/mol

75 g SiCl₄ x 1 mol/169.9 g = .441 mol SiCl₄

Moles to moles:

For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

.441 mol SiCl₄ = .441 mol SiO₂

Moles to mass:

molar mass SiO₂: 60.09 g/mol

.441 mol SiO₂ x 60.09g/mol = 26.5 g SiO₂

Theoretical yield:

actual/theoretical x 100

25.2 / 26.5 = .950 = 95.0% yield

Because the percentage yield of trial 2 is higher than that of trial 1, we know that the ratio of reactants in trial 2 is more efficient! We got a result closer to our theoretical yield.

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Sulfur is composed of three isotopes: 32S, 33S, and 34S. The atomic masses of these isotopes are given below. 32S: 31.97207 amu

elena-14-01-66 [18.8K]

Answer:

Abundance of 32S is 94.41%

Explanation:

The average atomic mass is defined as the sum of the atomic masses of each isotope times its abundance:

Average atomic mass = ∑ Atomic mass istope*Abundance

For the sulfur:

32.07amu = 31.97207X + 32.97146Y + 33.96786*0.0422 (1)

Where X is abundance of 32S and Y abundance of 33S

Also we can write:

1 = X + Y + 0.0422 (2)

0.9578 - X = Y

Because the sum of the abundances = 1

Replacing (2) in (1):

32.07amu = 31.97207X + 32.97146(0.9578 - X) + 33.96786*0.0422

32.07 = 31.97207X + 31.58006 - 32.97146X + 1.43344

-0.9435 = -0.99939X

0.9441 =X

In percentage, abundance of 32S is 94.41%

3 0

1 year ago

Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with

stepan [7]

Answer:

1) Net ionic equation :

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

2) 0.765 M is the molarity of the carbonic acid solution.

Explanation:

1) In aqueous carbonic acid , carbonate ions and hydrogen ion is present.:

$H_2CO_3(aq)\rightarrow 2H^+(aq)+CO_3^{2-}(aq)$ ..[1]

In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:

$KOH(aq)\rightarrow K^+(aq)+OH^{-}(aq)$ ..[2]

In aqueous potassium carbonate , potassium ions and carbonate ion is present.:

$K_2CO_3(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)$ ..[3]

$H_2CO_3(aq)+2KOH(aq)\rightarrow K_2CO_3(aq)+2H_2O(l)$

From one:[1] ,[2] and [3]:

$2H^+(aq)+CO_3^{2-}(aq)+2K^+(aq)+2OH^{-}(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)+H_2O(l)$

Cancelling common ions on both sides to get net ionic equation :

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2CO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\\M_1=?\\V_1=50.0 mL\\n_2=1\\M_2=3.840 M\\V_2=20.0 mL$

Putting values in above equation, we get:

$M_1=\frac{1\times 3.840 M\times 20.0 mL}{2\times 50.2 mL}=0.765 M$

0.765 M is the molarity of the carbonic acid solution.

6 0

2 years ago

Select the number of electrons each atom must gain or lose to have a full valence level. Use the periodic table if you need help

Lady bird [3.3K]

Calcium will loose one electron. Fluorine will gain one electron. Lithium will loose one electron. Argon will not loose any because it already has a full valence level. Aluminium will loose 3 electrons.

8 0

1 year ago

Read 2 more answers

Which compound(s) contain the most hydrogen atoms per molecule or formula unit? hydrogen selenide ammonium bromide strontium dih

igomit [66]

The answer is strontium dihydrogen phosphate

That is strontium dihydrogen phosphate, is the compound containing the most hydrogen atoms per molecule or formula unit.

The formulas of the following compounds are as follows:

Hydrogen selenide H₂Se

Ammonium bromide NH₃Br

Strontium dihydrogen phosphate Sr(H₂PO4)₂

Sodium bicarbonate NaHCO₃

As it can be seen from the formula of strontium dihydrogen phosphate Sr(H₂PO4)₂ , that it contains the most hydrogen atoms per molecule or formula unit.

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1 year ago

6. What is the molarity of 1.35 mol H2So4 in 245 mL of solution?

vovikov84 [41]

Answer:

5.51mol/L

Explanation:

Number of moles = 1.35moles

Volume of the solution = 245mL = 245*10^-3L = 0.245L

Molarity of a solution is the defined as the number of moles of a solute dissolved in 1L of the solution.

1.35 moles = 0.245L

X moles = 1L

X = (1.35 * 1) / 0.245

X = 5.51mol/L

The molarity of the solution is 5.51mol/L

7 0

1 year ago