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2 years ago

What is the mass of 22.4 L of H2O at STP?

1 answer:

3 0

Answer is: A. 18.02 g/mol.

At standard temperature and pressure 1 mol of gas occupied 22.4 liters:
V(H₂O) = 22.4 L; volume of water.
Vm = 22.4 L/mol; molar volume at STP.
n(H₂O) = V ÷ Vm.
n(H₂O) = 22.4 L ÷ 22.4 L/mol.
n(H₂O) = 1 mol; amount of substance (water).
M(H₂O) = Ar(O) + 2Ar(H) · g/mol.
M(H₂O) = 16 + 2 ·1.01 · g/mol.
M(H₂O) = 18.02 g/mol; molar mass of water.

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Help?

Ahat [919]

Answer:

$KClO_3$

Explanation:

Hello!

In this case, as we know the mass of the total sample, we can first compute the mass of oxygen:

$m_O=22.9g-7.33g-6.65g=8.92g$

Next, we compute the moles of each element:

$n_K=\frac{7.33g}{39.9g/mol}= 0.184mol\\\\n_{Cl}=\frac{6.65g}{35.45g/mol}=0.188mol \\\\n_O=\frac{8.92g}{16.00g/mol} =0.5575mol$

Now, we divide the moles by 0.184 moles, the fewest ones, to obtain:

$K=\frac{0.184}{0.184}=1.0 \\\\Cl=\frac{0.187}{0.184}=1.0\\\\O=\frac{0.5575}{0.184} =3.0$

Therefore, the empirical formula is:

$KClO_3$

Regards!

3 0

2 years ago

What 2 energy resources do not rely on photosynthesis

puteri [66]

The sun i believe is one and wind

6 0

3 years ago

Can someone check if i did this right. thank you !!

Tasya [4]

Answer:

noo

Explanation:

its basic

....................

6 0

2 years ago

Read 2 more answers

Which of the following 0.300 M solutions would contain the highest concentration of potassium ions?

mojhsa [17]

Here is a list of 0.300 M solutions that you can consider to select which has the highest concetration of potassium ions:

Select one:

a. potassium oxide

b. potassium phosphate

c. potassium hydrogen carbonate

d. potassium hypochlorite

e. potassium iodide

Answer:

Option b. potassium phosphate

Explanation:

First, you must write the chemical formula of each one of the potasssium compounds. Then, assume all the compounds ionize 100%.

Since all the solutions have the same concentrations, by assuming 100% ionization, the ionization equations will show which one produces the highest number of potassium ions, K⁺¹, and that is the solution with the highest concentration of such ions.

1. Chemical formulae

Compound Formula

a. potassium oxide K₂O

b. potassium phosphate K₃PO₄

c. potassium hydrogen carbonate KHPO₃

d. potassium hypochlorite KClO₄

e. potassium iodide KI

2. Dissociation (ionization):

a. potassium oxide, K₂O

K₂O does not dissociate but react with water to product KOH, then KOH dissociates, but there is not such thing as a 0.300 M solution of K₂O.

b. potassium phosphate:

K₃PO₄ → 3K³⁺ + PO₄¹⁻ 3 K⁺¹ ions per unit formula

c. potassium hydrogen carbonate

KHPO₃ → K⁺¹ + HPO₃⁻¹ 1 K⁺¹ ion per unit formula

d. potassium hypochlorite

KClO₄ → K⁺¹ + HClO₄⁻¹ 1 K⁺¹ ion per unit formula

e. potassium iodide

KI → K⁺¹ + I⁻¹ 1 K⁺¹ ion per unit formula.

Hence, the 0.300 M solution of potasssium phosphate produces the highest concentration of potassium ions.

6 0

3 years ago

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. CaCO 3 ( s ) + 2

denis23 [38]

Answer:

Mass of formed CaCl₂ is 14.9 g

Mass of CaCO₃ that remains after the reaction is 14.5 g

Explanation:

We determine the reaction:

CaCO₃ (s) + 2HCl (aq) ⟶ CaCl₂ (aq) + H₂O (l) + CO₂ (g)

We have the mass of both reactants, so we have to find out which is the limiting; firstly we convert the mass to moles

28 g . 1mol / 100g = 0.28 moles of CaCO₃

10 g . 1mol /36.45g =0.27 moles of HCl

Ratio is 1:2. Let's make a rule of three:

1 mol of carbonate needs 2 moles of acid, to react

Then, 0.28 moles must react with the double of moles, of acid.

Therefore the HCl is the limiting reactant, and the reagent in excess is the salt.

Ratio is 2:1. The rule of three will be:

2 moles of acid needs 1 mol of salt to react

Then, 0.27 moles of acid, will react with (0.27 .1 ) / 2 = 0.135 moles of salt

We have 0.28 moles of carbonate so (0.28 - 0.135) = 0.145 moles of salt will remain after the reaction is complete.

We convert the moles to mass → 0.145 mol . 100 g/1mol = 14.5 g

We said, that the limiting reagent was the HCl so we can work with the products, now. Ratio is 2:1

2 moles of HCl can produce 1 mol of CaCl₂

Then, 0.27 moles of HCl will produce (0.27 . 1)/ 2 = 0.135 moles of chloride

We convert the moles to mass → 0.135 mol . 110.98 g/1mol = 14.9 g

8 0

3 years ago