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marysya [2.9K]
3 years ago
13

Calcium hydroxide + hydrochloric acid → calcium chloride + water

Chemistry
2 answers:
Andrew [12]3 years ago
7 0

Answer : The correct balanced chemical reaction will be:

Ca(OH)_2(aq)+2HCl(aq)\rightarrow CaCl_2(s)+2H_2O(aq)

Explanation :

When calcium hydroxide react with hydrochloric acid then it react to give calcium chloride and water as a product.

This reaction is double-displacement reaction.

Double-displacement reaction : It is defined as the reaction in which the cation of two reactants molecule exchange their places to give two different products.

Thus, the correct balanced chemical reaction will be:

Ca(OH)_2(aq)+2HCl(aq)\rightarrow CaCl_2(s)+2H_2O(aq)

Lubov Fominskaja [6]3 years ago
5 0

Answer:

D

Explanation: iIt is a neutralization reaction and neutralization reaction is a reaction between and acid and base to give salt and water. In the question above, Ca(OH)2 is the base and HCl is the acid, Ca(Cl)2 is the salt and H2O.

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How many grams of oxygen gas (02) are needed to completely react with 9.30 moles
Luba_88 [7]

Answer:

223 g O₂

Explanation:

To find the mass of oxygen gas needed, you need to (1) convert moles Al to moles O₂ (via the mole-to-mole ratio from reaction coefficients) and then (2) convert moles O₂ to grams O₂ (via the molar mass). When writing your ratios/conversions, the desired unit should be in the numerator in order to allow for the cancellation of the previous unit. The final answer should have 3 sig figs because the given value (9.30 moles) has 3 sig figs.

4 Al + 3 O₂ ----> 2 Al₂O₃
^         ^

Molar Mass (O₂): 32.0 g/mol

9.3 moles Al          3 moles O₂              32.0 g
-------------------  x  ---------------------  x  --------------------  =  223 g O₂
                              4 moles Al               1 mole

6 0
2 years ago
You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

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