The mass of 63 ml sample : 79.38 g
<h3>Further explanation</h3>
Given
20 ml and 25.2 g of glycerol
Required
The mass of 63 ml sample
Solution
Density is the ratio of mass per unit volume
Density formula:
Density of glycerol :
= m : V
= 25.2 g : 20 ml
= 1.26 g/ml
Mass of 63 ml sample :
= density x volume
= 1.26 g/ml x 63 ml
= 79.38 g
And what are the compounds?
Answer:
You have to know what the three sates of matter are
Explanation:
The answer is: [B]: 1.89 * 10²³ .
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PV = nRT ; is the equation for "STP" conditions; that is, the "ideal gas equation" .
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i.e. when Pressure, "P" = 1.00 atm;
Temperature, "T" = 273 K;
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R = the ideal gas constant = ((0.08206 L-atm/K-mol)
n = number of moles;
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So, we plug in our known values:
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<span>(1.00atm) (7.02L) = ("n" mol) (0.08206 L-atm/K-mol) (273K); </span>
<span>_____________________________________________
</span>→ 7.02 L·atm = (? mol) (22.4 L·<span>atm/mol) .
</span>
(Note that t<span>he Molar Volume of a gas at STP is a constant using Avogadro's value of </span>22.4 L / mol. 1 mol of any ideal gas at STP occupies 22.4 L. An ideal gas takes the shape of its container)..
<span>_______________________________________________________
</span> → Divide EACH side of the equation by "(22.4 L·atm/mol)" ;
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→ 7.02 L·atm / = ("n" mol) (22.4 L·atm/mol) ;
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→ 7.02 L·atm / (22.4 L·atm/mol) =
[("n" mol) (22.4 L·atm/mol)]/(22.4 L·atm/mol);
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<span>to get:
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</span>→ n = <span>0.313 mol ;
</span>_____________________________________________________
Note: 1 mole = 1 mol = 6.022 * 10²³ molecules.
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→ So, 0.313 mol Cl₂ (g) molecules * [(6.022 * 10²³ molecules) / (1 mol)] =
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→ [(0.313) * (6.022 *10²³) ] molecules of Cl₂ (g) ;
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→ = 1.88 * 10²³ molecules of Cl₂ g ;
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→ which most closely corresponds with answer choice:
___________________________________________________________
[B]: 1.89 * 10²³ .
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Answer: Option (c) is the correct answer.
Explanation:
Boiling point is defined as the point at which vapor pressure of a liquid becomes equal to the atmospheric pressure.
Boiling point of water is 
.
Whereas when we heat one mole of a liquid at its boiling point without any change in temperature then the heat required to bring out change from liquid to vapor state is known as heat of vaporization.
Thus, we can conclude that the heat of vaporization of water is the amount of heat/energy required to convert 1 gram of liquid water at to 1 gram of steam at .
