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2 years ago

What volume of 0.500 M HNO3(aq) must completely react to neutralize 100.0 milliliters of 0.100 M KOH(aq)?

1 answer:

6 0

KOH+ HNO3--> KNO3+ H2O
From this balanced equation, we know that 1 mol HNO3= 1 mol KOH (keep in mind this because it will be used later).

We also know that 0.100 M KOH aqueous solution (soln)= 0.100 mol KOH/ 1 L of KOH soln (this one is based on the definition of molarity).

First, we should find the mole of KOH:
100.0 mL KOH soln* (1 L KOH soln/ 1,000 mL KOH soln)* (0.100 mol KOH/ 1L KOH soln)= 1.00*10^(-2) mol KOH.

Now, let's find the volume of HNO3 soln:
1.00*10^(-2) mol KOH* (1 mol HNO3/ 1 mol KOH)* (1 L HNO3 soln/ 0.500 mol HNO3)* (1,000 mL HNO3 soln/ 1 L HNO3 soln)= 20.0 mL HNO3 soln.

The final answer is (2) 20.0 mL.

Also, this problem can also be done by using dimensional analysis.

Hope this would help~

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At 700 K, Kp for the following equilibrium is (5.6 x 10-3) 2HgO(s)--> 2Hg(l) + O2(g) Suppose 51.2 g of mercury(II) oxide is p

Vesnalui [34]

Answer: Option (B) is the correct answer.

Explanation:

According to the given reaction equation, formula to calculate $\Delta n$ is as follows.

$\Delta n$ = coefficients of gaseous products - gaseous reactants

= 1 - 0

= 1

Also we know that,

$K_{p} = K_{c} \times (RT)^{\Delta n}$

$5.6 \times 10^{-3} = K_{c} \times (0.0821 \times 700)^{1}$

$K_{c} = 0.097 \times 10^{-3}$

For the equation, $2HgO(s) \rightarrow 2Hg(l) + O_{2}(g)$

Activity of solid and liquid = 1

As, $K_{p} = \frac{P^{2}_{Hg} \times P_{O_{2}}}{P^{2}_{HgO}}$

$5.6 \times 10^{-3} = P_{O_{2}}$

Hence, $P_{O_{2}}$ = 0.0056 atm

Thus, we can conclude that partial pressure of oxygen gas at equilibrium is 0.0056 atm.

5 0

2 years ago

How many liters of a 0.352 M solution of Ca(SO4) would contain 62.1 g of Ca(SO4)?

konstantin123 [22]

Answer:

1.3 L.

Explanation:

Molarity is the no. of moles of solute per 1.0 L of the solution.

M = (no. of moles of CaSO₄)/(Volume of the solution (L))

M = 0.352 M.

no. of moles of CaSO₄ = mass/molar mass = (62.1 g / 136.14 g/mol) = 0.456 mol,

Volume of the solution = ??? L.

∴ (0.352 M) = (0.456 mol)/(Volume of the solution (L))

∴ (Volume of the solution (L) = (0.456 mol)/(0.352 M) = 1.296 L ≅ 1.3 L.

4 0

2 years ago

What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter

Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole$

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

$44.5kJ/mol=\frac{q}{0.038mol}$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

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

$1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

4 0

2 years ago

What is the amount of heat released when 25g of water cools 12.5 degrees C ?

Over [174]

Water can't cool at a single temperature. It must start at a higher temperature, and drop to a lower temperature in order to cool. Unless we know the other temperature, there is no way to calculate the amount of thermal energy released.

4 0

2 years ago

Review the list of common titration errors. Determine whether each error would cause the calculation for moles of analyte to be

kakasveta [241]

Answer and Explanation:

A funnel is in the top of the buret and a beaker is positioned underneath the buret: This is correct and is necessary to fill the buret, but the funnel and the beaker has to be removed before the titration starts. The calculation for moles of analyte does not affect.

A solution is being poured from a bottle into the buret via the funnel: Using a funnel helps to fill the burette but it must be removed to filling the buret at 0.0 mL. In this case, the calculation for moles of analyte do not affect.

Adding titrant past the color change of the analyte solution: In this case, an excess of titrant is added, thus the calculation for moles of anality will be higher than it should be.

Recording the molarity of titrant as 0.1 M rather than its actual value of 0.01 M: In this case, the titrant is considered more concentrated than it is hence, the calculation for moles of anality will be higher than it should be.

Spilling some analyte out of the flask during the titration: The excess of titrant spilled out of the flask higher up the volume of titrant measured. Therefore, the calculation for moles of anality will be higher than it should be.

Starting the titration with air bubbles in the buret: The air inside the burette occupies measured volume, thus the volume of titrant measured will be higher than the real volume spilled in the flask. Hence the calculation for moles of anality will be higher than it should be.

Filling the buret above the 0.0 mL volume mark: Some volume of titrant will be spilled inside the flask but will no be measured since the buret measures the titrant below the 0.0mL mark, thus the calculation for moles of anality will be lower than it should be.

3 0

2 years ago