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3 years ago

Is it possible to convert between moles and mass?

Chemistry

2 answers:

lubasha [3.4K]3 years ago

5 0

Answer:

Yes, if you're talking about molar mass or grams

Explanation:

Hope this helps!

Valentin [98]3 years ago

5 0

Yes it is possible

Explain: have a great day! Hope this helps y’all!

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The equilibrium constant, K, for the following reaction is 1.80x102 at 698 K. 2HI(9) =H2(g) +129) An equilibrium mixture of the

Mrrafil [7]

Explanation:

The equilibrium constant of the reaction = $K_c=1.80\times 10^{-2}$

Initial concentration of HI = 0.311 M

After addition 0.174 mol of HI(g)

Concentration of HI added = $\frac{0.174 mol}{1 L}=0.174 M$

New concentration of HI = 0.311 M + 0.174 M = 0.485 M

$2HI\rightleftharpoons H_2+I_2$

Initial concentration:

0.311 M $4.71\times 10^{-2} M$ $4.71\times 10^{-2} M$

At equilibrium:

(0.485 M - x) $(4.71\times 10^{-2} M+x)$ $(4.71\times 10^{-2} M+x)$

$K_{eq}=\frac{[H_2][I_2]}{[HI]^2}$

$1.80\times 10^{-2}=\frac{(4.71\times 10^{-2} M+x)\times (4.71\times 10^{-2} M+x)}{ (0.485 M-x)^2}$

$\sqrt{1.80\times 10^{-2}}=\frac{(4.71\times 10^{-2} M+x)}{ (0.485 M-x)}$

$0.1342=\frac{(4.71\times 10^{-2} M+x)}{(0.485 M-x)}$

$0.065087 M-0.1342x=4.71\times 10^{-2} M+x$

$0.065087 M-4.71\times 10^{-2} M=1.1342x$

$x=\frac{0.017987 M}{1.3142}=0.01586 M$

Equilibrium concentrations:

$[HI]=0.485 M-x = 0.485 M - 0.01586 M= 0.46914 M$

$[H_2]=[I_2]=4.71\times 10^{-2} M+x=4.71\times 10^{-2} M+0.01586 M$

$[H_2]=[I_2]=0.06296 M$

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A chemist makes a solution of Ca(NO3)2 by dissolving 21.3 g Ca(NO3)2 in water to make 100.0 mL of solution. What is the concentr

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Answer:

[NO₃⁻ ] = 2.596 M

Explanation:

Ca(NO₃)₂ dissolves in water according to the following equation:

Ca(NO₃)₂ ⇒ Ca²⁺ + 2NO₃⁻

The moles of Ca(NO₃)₂ that dissolve is found as followed:

(21.3 g) / (164.10 g/mol) = 0.1298... mol

The number of NO₃⁻ ions are related to the above quantity by the molar ratio:

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