Answer: 7.79 grams of ethanol were put into the beaker.
Explanation:
To calculate the mass of ethanol, we use the equation:
Density of ethanol = 0.779 g/mL
Volume of water = 10.00 mL
Putting values in above equation, we get:
Thus 7.79 grams of ethanol were put into the beaker.
The answer for the following problem is mentioned below.
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
Explanation:
Given:
mass of methane() = 272 grams
pressure (P) = 250 k Pa =250×10^3 Pa
temperature(t) = 54°C =54 + 273 = 327 K
Also given:
R = 8.31JK-1 mol-1 ,
Molar mass of methane() = 16.0 grams
We know;
According to the ideal gas equation,
<u><em>P × V = n × R × T</em></u>
here,
n = m÷M
n =272 ÷ 16
<u><em>n = 17 moles</em></u>
Therefore,
250×10^3 × V = 17 × 8.31 × 327
V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )
V = 184.78 × 10^-3 liters
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
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Answer:
31.1°C
Explanation:
Given parameters:
Temperature = 88°F
The formula of the to convert is:
T°F = T°C - 32 / 1.8 =
Now input the parameters and solve;
T°F =
T°F = 31.1°C