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kipiarov [429]
4 years ago
7

A block of metal has a mass of 0.0694 kg. It displaces 3.6 mL of water. What is it’s density in g/cm^3

Chemistry
1 answer:
Ivan4 years ago
4 0

Answer:

19.28 g/cm^3 to the nearest hundredth.

Explanation:

The volume of water displaced = the volume of the metal.

density = mass / volume

0.0694 kg = 0.0694 * 1000

=  69.4 g.

Density = 69.4 / 3.6

= 19.28 g/cm^3.

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jek_recluse [69]

Answer:

Au. fifififiof

Explanation:

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Calculate the molar solubility of CaF2 in a 0.25 m solution of NaF(aq).
Kipish [7]

Answer:

6.4 × 10^-10 M

Explanation:

The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).

CaF2 will dissociate as follows:

CaF2 ⇌Ca2+ + 2F-

1 mole of Calcium ion (x)

2 moles of fluorine ion (2x)

NaF will also dissociate as follows:

NaF ⇌ Na+ + F-

Where Na+ = 0.25M

F- = 0.25M

The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M

Ksp = {Ca2+}{F-}^2

Ksp = {x}{0.25}^2

4.0 × 10^-11 = 0.25^2 × x

4.0 × 10^-11 = 0.0625x

x = 4.0 × 10^-11 ÷ 6.25 × 10^-2

x = 4/6.25 × 10^ (-11+2)

x = 0.64 × 10^-9

x = 6.4 × 10^-10

Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M

8 0
4 years ago
Write the two half-equations for the following redox reaction <br> 2H+ +2Br- +H2SO4= Br2 + SO2 +2H2O
Oduvanchick [21]

Explanation:

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7 0
3 years ago
What is the theoretical yield of a reaction?
mestny [16]
I think it's D, because theoretical yield is like, the yield you'd get if 100% of the reactants formed to make product. Well that's how I think of it, but it has something to do with limiting reagents and stuff. Sorry this isn't a really detailed explanation.
4 0
3 years ago
Complete the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (Ty
anastassius [24]

Answer :

(A) The dissociation reaction of HC_2H_3O_2 will be:

HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)

The equilibrium expression :

K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}

(B) The dissociation reaction of Co(H_2O)_6^{3+} will be:

Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)

The equilibrium expression :

K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}

(C) The dissociation reaction of CH_3NH_3^+ will be:

CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)

The equilibrium expression :

K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

(A) The dissociation reaction of HC_2H_3O_2 will be:

HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)

The equilibrium expression of HC_2H_3O_2 will be:

K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}

(B) The dissociation reaction of Co(H_2O)_6^{3+} will be:

Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)

The equilibrium expression of Co(H_2O)_6^{3+} will be:

K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}

(C) The dissociation reaction of CH_3NH_3^+ will be:

CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)

The equilibrium expression of CH_3NH_3^+ will be:

K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}

3 0
3 years ago
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