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svp [43]
2 years ago
11

How would one make a 2 M solution of a compound?

Chemistry
2 answers:
Sati [7]2 years ago
5 0

Answer:

B. By dissolving 2 moles of the compound in 1 L of water.

DiKsa [7]2 years ago
5 0

Answer:

<em>b</em>

Explanation:

<em>a pe x :)</em>

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What physical property of matter determines the earth ?
nikdorinn [45]

Answer:

Density is a physical property that is determined by dividing the mass of a given amount of a substance by its volume.

Explanation:

4 0
3 years ago
A glass flask whose volume is 1000 cm3 at a temperature of 0.800 C is completely filled with mercury at the same temperature. Wh
ira [324]

Answer: the coefficient of volume expansion of glass = 0.86/(1000 * 52) = 0.00001654 per degree.

Explanation:

Original volume of mercury = 1000 cm3.

The final volume of mercury considering its volume expansion quotient = 1000 + 1000*(1.8*10^-4 *52) = 1000 + 9.36 = 1009.36 cm^3

Considering the glass as a non expanding substance, the complete excess volume of 9.36 cm3 of mercury should have overflown the container, but due to the expansion of glass, the capacity of mercury containment increases and so a lesser amount of mercury flows out.

The amount of mercury that actually flowed out = 8.50 cm3.

So, the expansion of the glass container = 9.36-8.50 = 0.86 cm3.

Using the formula for coefficient of expansion,

coefficient of volume expansion of glass = 0.86/(1000 * 52) = 0.00001654 per degree.

5 0
3 years ago
In this experiment you will create solutions with different ratios of ethanol and water. What is the mole fraction of ethanol wh
Andrej [43]

Answer:

The mole fraction of ethanol is 0.6. A 10 mL volumetric pipette must be used for to measure the 10 mL of ethanol. The vessel should be clean and purged.

Explanation:

For calculating mole fraction of ethanol, the amount of moles ethanol must be calculated. Using ethanol density (0.778 g/mL), 10 mL of ethanol equals to 7.89 g of ethanol and in turn 0.17 moles of ethanol. The same way for calculate the amount of water moles (ethanol density=0.997 g/mL). 2 mL of water correspond to 0.11. The total moles are: 0.17+0.11=0.28. Mole fraction alcohol is: 0.17/0.28=0.6

5 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
How is 0.246¯¯¯¯ written as a fraction in simplest form?
KiRa [710]
0.246 =  \dfrac{246}{1000} = \dfrac{246 \div 2}{1000 \div 2} = \dfrac{123}{500}
4 0
2 years ago
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