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Nuetrik [128]
2 years ago
10

0.005 molar Ca(OH)2 solution is used in the titration.calculate the pH oh solution

Chemistry
1 answer:
Rus_ich [418]2 years ago
6 0

Ca(OH)₂: strong base

pOH = a . M

a = valence ( amount of OH⁻)

M = concentration

Ca(OH)₂ ⇒ Ca²⁺ + 2OH⁻ (2 valence)

so:

pOH = 2 x 0.005

pOH = 0.01

pH = 14 - 0.01 = 13.99

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For each of the salts below, match the salts that can be compared directly, using Ksp values, to estimate solubilities.
Anni [7]

Answer:

1 - Salts required  that can be compared directly, using Ksp values, to estimate solubilities for copper (II) sulfide are - CaSO_3 (Option b) ,BaCrO_4 (Option c) and CaS (Option d)

2 - Salts  that can be compared directly, using Ksp values, to estimate solubilities for zinc hydroxide are - Mg(OH)_2 (Option a)

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K_s_p values can be used to compare the solubility of salts that produce ions in a 1:1 ratio.

  1. CuS {copper (II)sulfide} dissociates with ions in ratio 1:1 .

Because the salts CaSO_3 , BaCrO_4 and CaS  have a 1:1 ion ratio, the solubilities of options b, c, and d can be compared to salt 1.

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4 0
3 years ago
How many grams of nan3 are required to produce 19.0 ft3 of nitrogen gas, about the size of an automotive air bag, if the gas has
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The  balanced chemical reaction is given as:

2NaN_{3}(s)\rightarrow 2Na(s)+3N_{2}(g)

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1 ft^{3}  = 28.3168

So, 19.0 ft^{3} = 19\times 28.3168 = 538.0192 L

Density is equal to the ratio of mass to the volume.

D=\frac{M}{V}

where, M = mass and V= volume (538.0192 L)

Substitute the value of density and volume in formula to get the value of mass.

1.25 g/L=\frac{M}{538.0192 L}

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Now, in 24.00 moles of nitrogen gas produced from= \frac{2 moles of sodium azide}{3 moles of nitrogen gas}\times 24.00 moles of nitrogen gas, moles of sodium azide.

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Mass of sodium azide in g  =  number of moles\times molar mass of sodium azide.

= 16 moles\times 65.00 g/mol

= 1040 g

Thus, mass of sodium azide which is required to produce 19.0 ft^{3} of nitrogen gas  = 1040 g





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ArbitrLikvidat [17]

Answer:

<h2><u><em>100 kcal of bond energy</em></u></h2>

<u><em></em></u>

8 0
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