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Nutka1998 [239]
3 years ago
10

Given the following information: benzoic acid = C6H5COOH hydrocyanic acid = HCN C6H5COOH is a stronger acid than HCN (1) Write t

he net ionic equation for the reaction that occurs when equal volumes of 0.050 M aqueous benzoic acid and sodium cyanide are mixed. It is not necessary to include states such as (aq) or (s).
Chemistry
1 answer:
Maurinko [17]3 years ago
5 0

Answer:

The net ionic equation is

C6H5COOH+ CN-= C6H5COO- + HCN

Explanation:

From the ionic equation

C6H5COOH + Na+ + CN- = C6H5COO- + Na+ + HCN

Only sodium is the spectator ion, so it cancels out, since C6H5COOH and HCN do not ionize completely they are left undissociated

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Could ethanol vapor collect in low spots?
Leviafan [203]

Answer:

Yes

Explanation:

Denatured ethanol fuel is a polar solvent, which is soluble in water. A

Polar solvent is a compound with a charge separation in chemical bonds, such as  alcohol, most acids, or ammonia. These have affinity with water and  will dissolve easily. Denatured fuel ethanol has a flash point of  -5 ° F and a vapor density of 1.5, indicating that it is heavier than air.

Consequently, ethanol vapors do not rise, similar to the gasoline vapors they are looking for  lower altitudes. The specific gravity of denatured fuel ethanol is 0.79, which  indicates that it is lighter than water and has a self-ignition temperature of 709 ° F and a  boiling point of 165-175 ° F. Like gasoline, the most denatured fuel,  the greatest danger of ethanol as an engine fuel component is its flammability.

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3 years ago
A gas occupies a volume of 1.00 L at 25.0°C. What volume will the gas occupy at 1.00 x10^2 °C?
Leno4ka [110]

Answer : The volume of gas occupy at 1.00\times 10^2^oC is, 1.25 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=1.00L\\T_1=25.0^oC=(25.0+273)K=298K\\V_2=?\\T_2=1.00\times 10^2^oC=((1.00\times 10^2)+273)K=373K

Putting values in above equation, we get:

\frac{1.00L}{298K}=\frac{V_2}{373K}\\\\V_2=1.25L

Therefore, the volume of gas occupy at 1.00\times 10^2^oC is, 1.25 L

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