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Kobotan [32]
3 years ago
9

What acid and what base would you choose to prepare the salt potassium perchlorate (kclo4)?

Chemistry
2 answers:
Nonamiya [84]3 years ago
8 0
HClO₄ + KOH → KClO₄ + H₂O

HClO₄ - perchloric acid
KOH - potassium hydroxide
dolphi86 [110]3 years ago
7 0

Answer : The perchloric acid and potassium hydroxide base is used to prepare the salt of potassium perchlorate, (KClO_4)

Explanation :

when the perchloric acid react with the potassium hydroxide as a base to form a salt of potassium perchlorate, (KClO_4)

The balanced chemical reaction will be,

HClO_4+KOH\rightarrow KClO_4+H_2O

By the stoichiometry, we can say that 1 mole of perchloric acid react with the 1 mole of potassium hydroxide base to give 1 mole of potassium perchlorate and 1 mole of water as a product.

Hence, the perchloric acid and potassium hydroxide base is used to prepare the salt of potassium perchlorate, (KClO_4)

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Alka-seltzer is an antacid that is dissolved in water before use. for an experiment, three tablets of alka-seltzer are dissolved
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Alka-seltzer in an antacid that contains a mixture of sodium bicarbonate and citric acid. When the tablet is dissolved in water, the reactants which are in solid form in tablet become aqueous and react with each other.

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NaHCO_{3}\ (aq) +  H_{3}C_{6}H_{5}O_{7} (aq)\rightarrow Na_{3}C_{6}H_{5}O_{7}(aq)+ CO_{2}(g)+H_{2}O (l)

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8 0
3 years ago
What is the [OH-] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 m
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Answer: The [OH^-] of a solution is 10^{-12} M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute

V_s = volume of solution in ml

moles of HCl = \frac{\text {given mass}}{\text {Molar mass}}=\frac{0.0912g}{36.5g/mol}=0.0025mol

Now put all the given values in the formula of molality, we get

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pH or pOH is the measure of acidity or alkalinity of a solution.

HCl\rightarrow H^++Cl^{-}

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1 mole of HCl gives 1 mole of H^+

Thus 0.01 moles of HCl gives =\frac{1}{1}\times 0.01=0.01 moles of H^+

Putting in the values:

[H^+][OH^-]=10^{-14}

[0.01][OH^-]=10^{-14}

[OH^-]=10^{-12}

Thus the [OH^-] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution is 10^{-12} M

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