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MatroZZZ [7]
3 years ago
4

A laboratory assistant needs to prepare 35.2 liters of hydrogen at 25.0°C and 101.3 kilopascals. This is the equation for the re

action:
2HCl + Ca → H2 + CaCl2

What volume of 2.3 M hydrochloric acid is required to produce this much gas? Use the ideal gas resource.

A. 0.625 L
B. 0.876 L
C. 1.18 L
D. 1.25 L

(Some chart that I don't understand)

Chemistry
2 answers:
Mars2501 [29]3 years ago
6 0

Answer:

Option A= 0.625 L

Explanation:

Given data:

Volume of hydrogen =35.2 L

Temperature = 25 °C = 25 + 273 = 298 K

Pressure = 101.3 Kpa = 0.99975327 atm

Molarity of HCl = 2.3 mol / L

Volume of HCl require = ?

Solution:

Formula:

PV = nRT

n = PV / RT

n = (0.99975327 atm × 35.27 L) / 0.0821 atm. L/ mol.K × 298 k

n = 35.19 atm . L / 24.4658 L. atm /mol

n = 1.438 mol

we know that molarity is equal to:

Molarity = number of moles / volume in liter

Volume in liter = Number of moles / Molarity

Volume in liter = 1.438 mol / 2.3 mol/ L

Volume in liter = 0.625 L

Ray Of Light [21]3 years ago
6 0

Answer:

D. 1.25

Explanation:

for plato users

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Answer:

% = 5.69%

Explanation:

To do this, we need to write the equations taking place here. First, this is a REDOX reaction where the hypoclorite and thiosulfate solution reacts. The balanced equations are:

ClO⁻ + 2I⁻ + 2H⁺ -------> Cl⁻ +  I₂ + H₂O

I₂ + 2S₂O₃²⁻ -----------> 2I⁻ + S₄O₆²⁻

We already have the required volume and concentration of the thiosulfate solution, so we can calculate the moles of thiosulfate. With this moles, we can calculate the moles of hypochlorite, then the mass and finally the %.

The moles of thiosulfate would be:

moles S₂O₃²⁻ = V * M

moles S₂O₃²⁻ = 0.01324 * 0.0732 = 9.69x10⁻⁴ moles

Now according to the above reactions, we can see that

moles I₂ = moles ClO⁻

and

moles I₂ / moles S₂O₃²⁻ = 1/2

Therefore, let's calculate the moles of ClO⁻:

moles ClO⁻ = 9.69x10⁻⁴ / 2 = 4.845x10⁻⁴ moles

Now, we can calculate the mass of these moles, using the molar mass of sodium hypochlorite which is 74.44 g/mol:

m = 74.44 * 4.845x10⁻⁴

m = 0.036 g

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% = 0.036 / 0.634 * 100

<h2>% = 5.69%</h2>
6 0
3 years ago
When a solution of ammonium sulfate is added to a
Artist 52 [7]

Answer:

Explanation:

(NH4)2SO4+Pb(NO3)2...............PbSO4+2NH4NO3

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3 years ago
What is the percent error if the measured value is 30.0 g and<br> the accepted value is 32.0 g?
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3 years ago
How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b)
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Explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide =\frac{100 g}{44 g/mol}=2.273 moles

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol = 2.273\times 10^{-3} kmol

2) 1 liter of ethyl alcohol of density 0.789 g/cm^3

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d = 0.789 g/cm^3

1 cm^3=1 mL

Mass of ethyl alcohol = m

m=d\times V=0.789 g/cm^3\times 1000 mL=789 g

Molar mass of  ethyl alcohol = 46 g/mol

Moles of ethyl alcohol = \frac{789 g}{46 g/mol}=17.152 mol

17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol

3) Volume of oxygen gas,V =1.5 m^3=1500 L

1 m^3= 1000 L

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

PV=nRT

n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol

0.01632 mol = 0.01632 × 0.001 kmol=1.632\times 10^{-5} kmol

4) Volume water in mixture = 1 L

Density of water =  1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L

Mass of water = 1000 g/L\times 1 L = 1000 g

Volume of alcohol = 2.5 L

Density of alcohol =  789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L

Mass of alcohol = 789 g/L\times 2.5 L = 1972.5 g

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

\frac{1000 g}{2972.5 g}\times 100=33.64\%

Mass percentage of alcohol :

\frac{1972.5 g}{2972.5 g}\times 100=66.36\%

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3 0
3 years ago
When solution A was mixed with solution B, the solution turned cloudy, the test tube did not feel warm, and there were no visibl
son4ous [18]

Answer:

Precipitation reaction

Explanation:

Given that solution A was mixed with solution B, the solution turned cloudy. The test is not warm and no bubbles visible. This means that the precipiate is formed.

The concept is when two colourless solutions react to form a cloudy precipitate that settles at bottom of a solution then the reaction is said to be a precipitation reaction.

An example can be the Reaction of Silver nitrate with common salt.

3 0
3 years ago
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