Answer:
132.17 g
Explanation:
The reaction given , in the question is -
CH₄ (g ) + 4 S ( g ) ---> CS₂ ( g ) + 2H₂S ( g )
From the reaction , 4 mole of S is required for the production of 1 mole of CS₂ .
since ,
Moles of CS₂ = given mass of CS₂ / Molecular weight of CS₂
Since ,
the Molecular weight of CS₂ = 76
Given , mass of CS₂ = 72.57 g
Moles of CS₂ = 72.57 / 76 = 0.95 mol
Since ,
The yield is 92.0 % .
Moles of S required = 4 * 0.95 mol / 0.92 = 4.13 moles
Mass of S required = 4.13 * 32 = 132.17 g .
Answer:
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The volume of the oxygen gas at standard temperature is 41.36 liters.
The given parameters;
- <em>initial volume of oxygen, V₁ = 45 L</em>
- <em>temperature of oxygen, T₁ = 24 ⁰C = 297 K</em>
- <em>standard temperature, T₂ = 0 ⁰C = 273 K</em>
<em />
The volume of the oxygen gas at standard temperature is determined by applying Charles law as shown below;
Thus, the volume of the oxygen gas at standard temperature is 41.36 L.
Learn more here:brainly.com/question/16927784
Answer:
The pressure of the gas would decrease
Explanation:
Answer:
a. The specific heat capacity of the gaseous ethanol is less than the specific heat capacity of liquid ethanol.
Explanation:
The heating curve is a curve that represents temperature (T) in the y-axis vs. added heat (Q) in the x-axis. The slope is T/Q = 1/C, where C is the heat capacity. Then, the higher the slope, the lower the heat capacity. For a constant mass, it can also represent the specific heat capacity (c).
Heats of vaporization and fusion cannot be calculated from these sections of the heating curve.
<em>Which statement below explains that?</em>
<em>a. The specific heat capacity of the gaseous ethanol is less than the specific heat capacity of liquid ethanol.</em> YES.
<em>b. The specific heat capacity of the gaseous ethanol is greater than the specific heat capacity of liquid ethanol.</em> NO.
<em>c. The heat of vaporization of ethanol is less than the heat of fusion of ethanol.</em> NO.
<em>d. The heat of vaporization of ethanol is greater than the heat of fusion of ethanol.</em> NO.
