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Amiraneli [1.4K]
3 years ago
6

How much energy would need to be input to completely boil 17.3 moles of ice starting at 0 degrees * C and going to 100°C?

Chemistry
1 answer:
kolezko [41]3 years ago
4 0

Answer:

Q = 934.911\,kJ

Explanation:

The energy needed to boil 17.3 moles of ice is:

Q = (17.3\,mol)\cdot \left[\left(40.5\,\frac{kJ}{mol} \right)+ \left(\frac{4.186}{1000}\,\frac{kJ}{g\cdot ^{\circ}C}  \right)\cdot \left(\frac{18.015\,g }{1\,mol} \right)\cdot (100^{\circ}C-0^{\circ}C)+\left(6\,\frac{kJ}{mol}\right)\right]Q = 934.911\,kJ

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it is a base

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Can someone please help me with this question also explain the answers I am so confused thank you.
Archy [21]

The theoretical yield of H₂S is 13.5 g.

The percent yield is 75.5 %.

<h3>What is the theoretical yield of H₂S from the reaction?</h3>

The equation of the reaction is given below:

  • FeS + 2 HCl → FeCl₂+ H₂S

Moles of FeS reacting = mass/molar mass

Molar mass of FeS = 88 g/mol

Moles of FeS reacting = 35/88 = 0.398 moles

Moles of H₂S produced = 0.398 moles

Molar mass of H₂S = 34 g/mol

Mass of H₂S produced = 0.398 * 34 = 13.5 g

Theoretical yield of H₂S is 13.5 g.

  • Percent yield = actual yield/theoretical yield * 100%

Actual yield of H₂S = 10.2 g

Percent yield = 10.2/13.5 * 100%

Percent yield = 75.5 %

In conclusion, the actual yield is less than the theoretical yield.

Learn more about percent yield at: brainly.com/question/8638404

#SPJ1

4 0
1 year ago
After one half life ___% of the atoms will change to a stable
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2 years ago
. How many milliliters of 0.20 M HCl are needed to exactly neutralize 40. milliliters of 0.40 M KOH
Anuta_ua [19.1K]

Answer:

V_{HCl}=80mL

Explanation:

Hello,

In this case, for the given reactants we identify the following chemical reaction:

KOH+HCl\rightarrow KCl+H_2O

Thus, we evidence a 1:1 molar ratio between KOH and HCl, therefore, for the complete neutralization we have equal number of moles, that in terms of molarities and volumes become:

n_{HCl}=n_{KOH}\\\\M_{HCl}V_{HCl}=M_{KOH}V_{KOH}

Hence, we compute the volume of HCl as shown below:

V_{HCl}=\frac{M_{KOH}V_{KOH}}{M_{HCl}} =\frac{0.40M*40mL}{0.20M} \\\\V_{HCl}=80mL

Best regards.

5 0
3 years ago
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