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noname [10]
3 years ago
14

What is the volume of 500g of CO2?

Chemistry
2 answers:
taurus [48]3 years ago
6 0

Weighs 0.001836 gram per cubic centimeter or 1.836 kilogram per cubic meter

Airida [17]3 years ago
3 0

Weighs 0.001836 gram per cubic centimeter or 1.836 kilogram per cubic meter


Try to see if this helps

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5 0
4 years ago
Determine the concentration of nh3(aq) that is required to dissolve 743 mg of agcl(s) in 100.0 ml of solution. the ksp of agcl i
AnnyKZ [126]

Answer:

1.1 M

Explanation:

The dissociation of AgCl_{(s)} is as follows:

AgCl \leftrightharpoons Ag^+_{(aq)}+Cl^-_{(aq)}

Given Value for K_{sp} = 1.77*10^{-10}

The equation for the reaction for the formation of complex ion Ag(NH_3)^+_2 is :

Ag^+_{(aq)}    +     2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)}

The value of K_f = 1.6*10^7

If we combine both equation and find the overall equilibrium constant will be:

AgCl \leftrightharpoons Ag^+_{(aq)}+Cl^-_{(aq)}

Ag^+_{(aq)}    +     2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)}

<u>                                                                                                      </u>

AgCl_{(s)}+2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)} + Cl^-_{(aq)}

                                 K = (1.77*10^{-10})(1.6*10^7)

                                  K = 0.00283

If [NH_3] = x M

The solubility of  AgCl_{(s)} in the NH_3 solution will be:

x = 743*10^{-3}g * \frac{mol AgCl}{143.32g}*\frac{1}{0.1000L}

x = 0.0518 M

Constructing an ICE Table; we have :

                             AgCl_{(s)} + 2 NH_3{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)} + Cl^-_{(aq)}

Initial  (M)                                     x                   0                      0

Change  (M)                       -2 (0.0518)      + 0.0518          + 0.0518

Equilibrium (M)                    x - 0.1156          0.0518             0.0518

Equilibrium constant;

(K) = \frac{[Ag(NH_3)_2^+][Cl^-]}{[NH_3]^2}

0.00283 = \frac{(0.0518)^2}{(x-0.1156)^2}

0.00283 = (\frac{0.0518}{x-0.1156})^2

x = 0.1156 + \sqrt{\frac{0.0518^2}{0.00283} }

x = [NH₃] =  1.089 M

[NH₃] ≅ 1.1 M                

8 0
3 years ago
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