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3 years ago

If NaCl has a mass of 3.2g, what is the volume of chlorine gas at STP?

Chemistry

2 answers:

Shalnov [3]3 years ago

5 0

Hey there!

Molar mass NaCl = 58.44 g/mol

58.44 g ----------------- 22.4 ( at STP )

3.2 g -------------------- Volume ??

Volume = ( 3.2 x 22.4 ) / 58.44

Volume = 71.68 / 58.44

Volume = 1.226 L

Hope this helps!

a_sh-v [17]3 years ago

4 0

Answer:

V = 0.56 L

Explanation:

Given data:

Mass of NaCl = 3.2 g

Volume of chlorine gas = ?

Pressure and temperature = standard

Solution:

Chemical equation:

2NaCl → 2Na + Cl₂

Number of moles of NaCl:

Number of moles = mass/molar mass

Number of moles = 3.2 g/ 58.44 g/mol

Number of moles = 0.05 mol

Now we will compare the moles of NaCl with chlorine;

NaCl : Cl₂

2 : 1

0.05 : 1/2×0.05 = 0.025 mol

Volume of chlorine:

PV = nRT

V = nRT/P

V = 0.025 mol × 0.0821 atm.L/mol.K × 273 K / 1 atm

V = 0.56 L

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3 years ago

A solution was made by mixing 60.5 g of glucose (C6H12O6) with enough water to make 325 ml of solution. What is the percent (mas

IgorC [24]

mass percent concentration = 15.7 %

molar concentration of glucose solution 1.03 M

Explanation:

To calculate the mass percent concentration of the solution we use the following formula:

concentration = (solute mass / solution mass) × 100

solute mass = 60.5 g

solution mass = solute mass + water mass

solution mass = 60.5 + 325 = 385.5 g (I used the assumption that the solution have a density of 1 g/mL)

concentration = (60.5 / 385.5) × 100 = 15.7 %

Now to calculate the molar concentration (molarity) of the solution we use the following formula:

molar concentration = number of moles / volume (L)

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number of moles of glucose = 60.5 / 180 = 0.336 moles

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4 years ago

A buffer solution is prepared by placing 5.86 grams of sodium nitrite and 32.6 mL of a 4.90 M nitrous acid solution into a 500.0

N76 [4]

This question is incomplete, the complete question is;

A buffer solution is prepared by placing 5.86 grams of sodium nitrite and 32.6 mL of a 4.90 M nitrous acid solution into a 500.0 mL volumetric flask and diluting to the calibration mark. If 10.97 mL of a 1.63 M solution of potassium hydroxide is added to the buffer, what is the final pH? The Ka for nitrous acid = 4.6 × 10⁻⁴.

Answer:

the final pH is 3.187

Explanation:

Given the data in the question;

Initial moles of HNO2 = 32.6/1000 × 4.90 = 0.15974 mol

Initial moles of NO2- = mass/molar mass = 5.86/68.995 = 0.0849336 mol

Moles of KOH added = 10.97/1000 × 1.63 = 0.0178811 mol

HN02 + KOH → NO2- + H2O

moles of HNO2 = 0.15974 - 0.0178811 = 0.1418589 mol

Moles of NO2- = 0.0849336 + 0.0178811 = 0.1028147 mol

Now,

pH = pka + log( [NO2-]/[HNO2])

pH = -log ka + log( moles of NO2- / moles of HNO2 )

we substitute

pH = -log( 4.6 × 10⁻⁴ ) + log( 0.1028147 / 0.1418589 )

pH = -log( 4.6 × 10⁻⁴ ) + log( 0.724767 )

pH = 3.337242 + (-0.1398 )

pH = 3.187

Therefore, the final pH is 3.187

8 0

3 years ago