Why is potassium nitrate classified as an electrolyte?.

How many grams of KCl is needed to make .75 L of a 1 M solution of KCl?

Zielflug [23.3K]

Answer:

The answer to your question is 25.9 g of KCl

Explanation:

Data

Grams of KCl = ?

Volume = 0.75 l

Molarity = 1 M

Formula

$Molarity = \frac{number of moles}{volume}$

Solve for number of moles

$Number of moles = Molarity x volume$

Substitution

Number of moles = 1 x 0.75

Simplification

Number of moles = 0.75 moles

Molecular mass KCl = 39 + 35.5 = 34.5

Use proportions to find the grams of KCl

34.5 g of KCl ---------------- 1 mol

x ---------------- 0.75 moles

x = (0.75 x 34.5) / 1

x = 25.9 g of KCl

3 0

3 years ago

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

8 0

3 years ago

How does acid affect limestone?

Andrej [43]

Answer:

It reacts by fizzing and wearing away/dissolving the rock.

Explanation:

7 0

3 years ago

How many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 1.38 g of fe2s3 if the percent yield

ahrayia [7]

Answer: 100 ml


Explanation:

1) Convert 1.38 g of Fe₂S₃ into number of moles, n

i) Formula: n = mass in grass / molar mass

ii) molar mass of Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol


iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of Fe₂S₃


2) Use the percent yield to calculate the theoretical amount:


65% = 0.65 = actual yield/ theoretical yield =>

theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol Fe₂S₃

3) Chemical equation:


 3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)

4) Stoichiometrical mole ratios:


3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl

5) Proportionality:

2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃

=> x = 0.020 mol FeCl₃

6) convert 0.020 mol to volume


i) Molarity formula: M = n / V


ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml

3 0

3 years ago

Read 2 more answers

Solid carbon reacts with oxygen gas to form carbon dioxide gas . Write a balanced chemical equation for this reaction

Nikitich [7]

Solid carbon reacts with oxygen gas to form carbon dioxide gas.

The balanced chemical equation for this reaction:

$C (s) + O_{2}(g) ---> CO_{2}(g)$

7 0

4 years ago

Read 2 more answers