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3 years ago

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How many grams of a stock solution that is 87.0 percent H2SO4 by mass would be needed to make 200 grams of a 35.5 percent by mas

s solution? Show all of the work needed to solve this problem

1 answer:

Jet001 [13]3 years ago

5 0

Answer: Use c1(v1) = c2(v2) to solve<span> that </span>problem. 87%(200) = (35.5<span>%)v2 v2 ≈ 490.5. I hope this helps! Source(s)</span>

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What is the mass of a 1.68-l sample of a liquid that has a density of 0.921g/ml?

Gwar [14]

Hey there!

Volume in mL :

1.68 L * 1000 => 1680 mL

Density = 0.921 g/mL

Therefore:

Mass = density * Volume

Mass = 0.921 * 1680

Mass = 1547.28 g

7 0

3 years ago

What is the mass, in grams, of 50.0L of N2 at STP?

vodka [1.7K]

Hey there!:

Molar mass N₂ = 28.0134 g/mol

28.0134 g ------------------- 22.4 L (at STP )

mass N₂ -------------------- 50.0 L

mass N₂ = 50.0 x 28.0134 / 22.4

mass N₂ = 1400.67 / 22.4

mass N₂ = 62.529 g

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3 years ago

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th

MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

$\frac{K_1}{K_2}=?$

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

where,

$K_2$ = rate constant reaction -1

$K_1$ = rate constant reaction -2

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = $25^oC=273+25=298 K$

Now put all the given values in this formula, we get

$\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340$

0.4284 is the ratio of the rate constants.

7 0

3 years ago

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Lana71 [14]

Explanation:

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2 years ago

Read 2 more answers

(6 points) Calculate the maximum number of moles and grams of H2S that can form when 158 g of aluminum sulfide reacts with 131 g

Schach [20]

Answer:107.1 g, 124.1 g

Explanation:

The equation of the reaction is;

Al2S3(s) + 6H20(l) ----> 2Al(OH)3(s) + 3H2S(g)

Hence;

For Al2S3

Number of moles= reacting mass/molar mass

Number of moles = 158g/150gmol-1 =1.05 moles

If 1 mole of Al2S3 yields 3 moles of H2S

1.05 moles of Al2S will yield

1.05 × 3/1 = 3.15 moles

Mass of H2S = 3.15moles × 34 gmol-1 = 107.1 g

For water

Number of moles of water = 131g/18gmol-1= 7.3 moles

6 moles of water yields 3 moles of H2S

7.3 moles of water will yield 7.3 × 3/6 = 3.65 moles of H2S

3.65 moles × 34 gmol-1 =124.1 g

5 0

3 years ago