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Naya [18.7K]
3 years ago
12

Calculate the mass percent composition of sulfur in Al2(SO4)3.

Chemistry
1 answer:
Anastaziya [24]3 years ago
6 0
The molar mass of aluminum sulftae is 342.14 g/mol.

Since the subscript shows that there are 3 sulfurs within the substance, the total mass of sulfur is 96.21g/mol

Now take the mass of the sulfur and divide it by the molar mass of aluminum sulfate, then multiply by 100:
(96.21/342.15)(100) = 28.1% mass composition of sulfate
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The molar mass (atomic weight ) of sodium is 23.0 grams/mole and the molar mass of sodium azide, NaN3 , is the mass of sodium, 23.0 gram/mole added to the molar mass of three atoms of nitrogen (14.0 x 3 = 42 gram/mole) which equals 65.0 grams/mole. The percentage of sodium is 23.0 /65.0 x 100 % = 35 %
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What is the overall voltage for the nonspontaneous redox reaction involving
Ann [662]

Answer:

Answer D => E°(Mg°/Cu⁺²) = 0.34 + 2.37 = 2.71v

Explanation:

(Oxidation) =>  Mg°(s) => Mg⁺²(aq) + 2e⁻    E°(Mg°/Mg⁺²) = -2.37 v

(Reduction) => Cu⁺²(aq) + 2e⁻ => Cu°(s)     E°(Cu⁺²/Cu°) = +0.34 v

________________________________________________

Net Rxn => Mg°(s) + Cu⁺²(aq) => Mg⁺²(aq) + Cu°(s)

Std Cell Potential (25°C/1Atm) = E°(Redn) = E°(Oxidn) = +0.34v - (-2.37v)

= 0.34v + 2.37v = 2.72v

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3 years ago
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Calculate the mass in grams in nine molecules of CH3COOH? Please show how you got your answer.
WINSTONCH [101]
M CH₃COOH: 12u×2 + 1u×4 + 16u×2 =<u> 60u</u>

m 9CH₃COOH: 60u×9 = <u>540u</u>

<em>(1u ≈ 1,66·10⁻²⁴g)</em>
-----------------------------
1u ------- <span>1,66·10⁻²⁴g
540u ---- X
X = 540</span>×<span>1,66·10⁻²⁴g
<u>X = 896,4</u></span><span><u>·10⁻²⁴g


</u>
</span>
7 0
3 years ago
A weather balloon is filled with helium that occupies a volume of 5.57 104 L at 0.995 atm and 32.0°C. After it is released, it r
Alchen [17]

6.52 × 10⁴ L. (3 sig. fig.)

<h3>Explanation</h3>

Helium is a noble gas. The interaction between two helium molecules is rather weak, which makes the gas rather "ideal."

Consider the ideal gas law:

P\cdot V = n\cdot R\cdot T,

where

  • P is the pressure of the gas,
  • V is the volume of the gas,
  • n is the number of gas particles in the gas,
  • R is the ideal gas constant, and
  • T is the absolute temperature of the gas in degrees Kelvins.

The question is asking for the final volume V of the gas. Rearrange the ideal gas equation for volume:

V = \dfrac{n \cdot R \cdot T}{P}.

Both the temperature of the gas, T, and the pressure on the gas changed in this process. To find the new volume of the gas, change one variable at a time.

Start with the absolute temperature of the gas:

  • T_0 = (32.0 + 273.15) \;\text{K} = 305.15\;\text{K},
  • T_1 = (-14.5 + 273.15) \;\text{K} = 258.65\;\text{K}.

The volume of the gas is proportional to its temperature if both n and P stay constant.

  • n won't change unless the balloon leaks, and
  • consider P to be constant, for calculations that include T.

V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}.

Now, keep the temperature at T_1 =258.65\;\text{K} and change the pressure on the gas:

  • P_1 = 0.995\;\text{atm},
  • P_2 = 0.720\;\text{atm}.

The volume of the gas is proportional to the reciprocal of its absolute temperature \dfrac{1}{T} if both n and T stays constant. In other words,

V_2 = V_1 \cdot\dfrac{\dfrac{1}{P_2}}{\dfrac{1}{P_1}} = V_1\cdot\dfrac{P_1}{P_2} = 4.72122\times 10^{4}\;\text{L}\times\dfrac{0.995\;\text{atm}}{0.720\;\text{atm}}=6.52\times 10^{4}\;\text{L}

(3 sig. fig. as in the question.).

See if you get the same result if you hold T constant, change P, and then move on to change T.

6 0
2 years ago
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