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2 years ago

C5H12() + 502(9)=

Chemistry

1 answer:

Mariulka [41]2 years ago

3 0

The correct answer is B

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a scientist uses 68 grams of CaCo3 to prepare 1.5 liters of solution. what is the molarity of this solution?

victus00 [196]

Answer: 0.4533mol/L

Explanation:

Molar Mass of CaCO3 = 40+12+(16x3) = 40+12+48 = 100g/mol

68g of CaCO3 dissolves in 1.5L of solution.

Xg of CaCO3 will dissolve in 1L i.e

Xg of CaCO3 = 68/1.5 = 45.33g/L

Molarity = Mass conc.(g/L) / molar Mass

Molarity = 45.33/100 = 0.4533mol/L

7 0

2 years ago

Helium gas diffuses 4 times as fast as an unknown gas. What is relative molecular mass of the gas

Inessa05 [86]

The relative molecular mass of the gas : 64 g/mol

<h3>Further explanation</h3>

Given

Helium rate = 4x an unknown gas

Required

The relative molecular mass of the gas

Solution

Graham's Law

$\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

r₁=4 x r₂

r₁ = Helium rate

r₂ = unknown gas rate

M₁= relative molecular mass of Helium = 4 g/mol

M₂ = relative molecular mass of the gas

Input the value :

$\tt \dfrac{4r_2}{r_2}=\sqrt{\dfrac{M_2}{4} }\\\\16=\dfrac{M_2}{4}\\\\M_2=64~g/mol$

7 0

2 years ago

Last question, I promise!!!!!! ... :>

VikaD [51]

Answer:

Which question?

Explanation:

3 0

3 years ago

Read 2 more answers

Near the surface of a liquid, fast-moving particles can break free and become a gas.

Nookie1986 [14]

This statement is True! Lets think about it... When water boils, the water doesnt evaporate from the bottom of the bowl it evaporates from the top!
=)

6 0

2 years ago

The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Calculate the equilibrium concentration of H3O+ in the s

raketka [301]

Answer : The equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

Explanation :

The dissociation of acid reaction is:

$C_6H_5COOH+H_2O\rightarrow H_3O^++C_6H_5COO^-$

Initial conc. c 0 0

At eqm. c-x x x

Given:

c = $7.0\times 10^{-2}M$

$K_a=6.3\times 10^{-5}$

The expression of dissociation constant of acid is:

$K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}$

$K_a=\frac{(x)\times (x)}{(c-x)}$

Now put all the given values in this expression, we get:

$6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}$

$x=2.1\times 10^{-3}M$

Thus, the equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

4 0

2 years ago