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amm1812
3 years ago
11

25 g of ethyl alcohol is dissolved in 100 ml of water (density = 0.99993 at 20 oc). what is the % w/w for the ethyl alcohol in t

he resulting solution?
Chemistry
1 answer:
vitfil [10]3 years ago
5 0
<span>Percentage by mass is the amount in mass of a component in a mixture per 100 unit of mass of the total mixture. Percentage by mass is the same as  %w/w. We can determine this by dividing the mass of the solute with the total mass of the mixture. However, from the problem statement, we are given the volume of the water so there is a need to convert this value to mass by using the density of water. We calculate as follows:

Mass of solution = 100 mL (0.99993 g/mL) water + 25 g EtOH
Mass of solution = 124.993 g solution

%w/w = 25 g / 124.993 g x100
%w/w = 20% of EtOH</span>
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In an electron-dot structure of an element, the dots are used to represent ________.
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A gas is contained in a cylinder with a volume of 2.9 L at a temperature of 32.7oC and a pressure of 645.3 torr. The gas is then
stealth61 [152]

Answer: 41 atm

Explanation:

Given that:

Original Volume of gas V1 = 2.9L

Temperature T1 = 32.7°C

Convert Celsius to Kelvin

(32.7°C + 273 = 305.7K)

Pressure P1 = 645.3 torr

New Volume V2 = 0.23 L

New temperature T2 = 894.7°C

Convert Celsius to Kelvin

(894.7°C + 273 = 1167.7K)

New pressure = ?

Then, apply the combined gas equation

(P1V1)/T1 = (P2V2)/T2

(645.3 torr x 2.9L)/305.7K = (P2 x 0.23L)/1167.7K

1871.37 / 305.7 = 0.23P2 / 1167.7

To get P2, Cross multiply

1871.37 x 1167.7 = 305.7 x 0.23P2

2185198.749 = 70.311P2

Divide both sides by 70.311

2185198.749/70.311 = 70.311P2/70.311

31079.045 torr = P2

Now, convert pressure in torr to atmosphere

Since 760 torr = 1 atm

31079.045 torr = Z

cross multiply

760 torr x Z = 31079.045 torr x 1 atm

Z = 31079.045 torr / 760 torr

Z = 40.89 atm (Round to the nearest whole number as 41 atm)

Thus, new pressure of gas is 41 atm

3 0
3 years ago
The heat of reaction for the combustion of propane is –2,045 kJ. This reaction is: C3H8(g) +502 (g) 3CO2 (g) + 4H2O (g). Determi
Ede4ka [16]

Answer:

\Delta _fH_{C_3H_8}=-102.7kJ/mol

Explanation:

Hello,

In this case, we can consider that the given heat of combustion is indeed the heat of reaction since it corresponds to the combustion of propane, which is computed by using the heat formation of all the involved species as shown below:

\Delta _cH=3*\Delta _fH_{CO_2}+4*\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-5*\Delta _fH_{O_2}

Thus, since the heat of formation of gaseous carbon dioxide is -393.5 kJ/mol, water -241.8 kJ/mol and oxygen 0 kJ/mol, the heat of formation of propane is:

\Delta _fH_{C_3H_8}=3*\Delta _fH_{CO_2}+4*\Delta _fH_{H_2O}-5*\Delta _fH_{O_2}-\Delta _cH\\\\\Delta _fH_{C_3H_8}=3*(-393.5)+4*(-241.8)-5*0-(-2045)\\\\\Delta _fH_{C_3H_8}=-102.7kJ/mol

Best regards.

8 0
3 years ago
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