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kicyunya [14]
3 years ago
11

Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. neutraliz

ation reaction: Suppose 0.250 L of 0.450 M H2SO4 is mixed with 0.200 L of 0.240 M KOH . What concentration of sulfuric acid remains after neutralization
Chemistry
1 answer:
madreJ [45]3 years ago
4 0

Answer:

0.197 M

Explanation:

The reaction equation is:

H2SO4(aq) +2KOH(aq) ----> K2SO4(aq) + 2H2O(l)

number of moles of H2SO4 = 0.25 L * 0.45 M = 0.1125 moles

number of moles of KOH = 0.2 L * 0.24 M = 0.048 moles

since H2SO4 is the reactant in excess;

2 moles of KOH reacts with 1 mole of H2SO4

0.048 moles of KOH reacts with 0.048 * 1/2 = 0.024 moles of H2SO4

Amount of excess H2SO4 left unreacted = 0.1125 - 0.024 = 0.0885 moles

Total volume = 0.25 L+ 0.2 L = 0.45 L

concentration of H2SO4 = 0.0885/0.45 = 0.197 M

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Jlenok [28]

Answer: The new pressure of the gas, assuming that no gas escaped during the experiment is 0.470 atm

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 0.732 atm

P_2 = final pressure of gas = ?

V_1 = initial volume of gas = 6.87 L

V_2 = final volume of gas = 9.22 L

T_1 = initial temperature of gas = 95^oC=273+95=368K

T_2 = final temperature of gas = 44^oC=273+44=317K

Now put all the given values in the above equation, we get:

\frac{0.732\times 6.87}{368}=\frac{P_2\times 9.22}{317}

P_2=0.470atm

Thus the new pressure of the gas, assuming that no gas escaped during the experiment is 0.470 atm

7 0
4 years ago
A gas occupies a volume of 202 ml at a pressure of 505 torr. To what pressure must the gas be subject in order to change the vol
UNO [17]

Answer:

1569 torr

Explanation:

Assuming ideal behaviour and constant temperature, we can solve this problem by using <em>Boyle's law</em>, which states:

  • V₁P₁ = V₂P₂

Where in this case:

  • V₁ = 202 mL
  • P₁ = 505 torr
  • V₂ = 65.0 mL
  • P₂ = ?

We <u>input the data given by the problem</u>:

  • 202 mL * 505 torr = 65.0 mL * P₂

And <u>solve for P₂</u>:

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6 0
3 years ago
Rachel has a sample containing 2 moles of carbon. How many atoms of carbon are in this sample?
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Answer is: c. 1.204 × 10²⁴ atoms of carbon.

n(C) = 2 mol; amount of substance of carbon.

Na = 6.02·10²³ 1/mol; Avogadro constant (the number of constituent particles, in this example atoms, that are contained in the amount of substance given by one mole).

N(C) = n(C) · Na.

N(C) = 2 mol · 6.02·10²³ 1/mol.

N(C) = 12.04·10²³ = 1.204·10²⁴; number of carbon atoms in a sample.


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Answer:

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