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3 years ago

6

what volume of carbon 4 oxide measured at STP will produce when 21 gram of sodium hydrogen carbonate for is completely decompose

d accor. ding to the equation. (Na=23. H=1 C=12 O=16) molar vol =22.4dmcube. pleeeeeeeeeeese help,

1 answer:

Arlecino [84]3 years ago

8 0

The volume of CO₂ at STP = 2.8 L

<h3>Further explanation</h3>

Given

21 gram of sodium hydrogen carbonate-NaHCO₃

Required

Volume of CO₂

Solution

The decomposition of sodium bicarbonate into sodium carbonate, carbon dioxide, and water :

<em> 2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)</em>

mol of NaHCO₃ :

= mass : MW NaHCO₃

= 21 g : 84 g/mol

= 0.25

From the equation, mol ratio of NaHCO₃(s) :CO₂(g) = 2 : 1, so mol CO₂ :

= 1/2 x mol NaHCO₃

= 1/2 x 0.25

= 0.125

At STP, 1 mol gas = 22.4 L, so for 0.125 mol :

= 0.125 x 22.4 L

= 2.8 L

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2 years ago

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.8

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Complete Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?

Answer:

The pK_a value is $pK_a =7.82$

Explanation:

From the question we are told

The volume of base is $V_B = 26.mL = 0.0260L$

The pH of solution is $pH = 7.82$

The concentration of the acid is $C_A = 0.1M$

From the pH we can see that the titration is between a strong base and a weak acid

Let assume that the the volume of acid is $V_A = 18mL= 0.018L$

Generally the concentration of base

$C_B = \frac{C_AV_A}{C_B}$

Substituting value

$C_B = \frac{0.1 * 0.01800}{0.0260}$

$C_B= 0.0692M$

When 13mL of the base is added a buffer is formed

The chemical equation of the reaction is

$HA_{(aq)} + OH^-_{(aq)} --------> A^{+}_{(aq)} + H_2 O_{(l)}$

Now before the reaction the number of mole of base is

$No \ of \ moles[N_B] = C_B * V_B$

Substituting value

$N_B = 0.01300 * 0.0692$

$= 0.0009 \ moles$

Now before the reaction the number of mole of acid is

$No \ of \ moles = C_B * V_B$

Substituting value

$N_A = 0.01800 *0.1$

$= 0.001800 \ moles$

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this mathematically represented as

$N_B ' = N_B - N_B = 0$

The number of moles of acid is

$N_A ' = N_A - N_B$

$= 0.0009\ moles$

The pH of this reaction can be mathematically represented as

$pH = pK_a + log \frac{[base]}{[acid]}$

Substituting values

$7.82 = pK_a +log \frac{0.0009}{0.0009}$

$pK_a =7.82$

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2 years ago

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TiliK225 [7]

Answer:

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Explanation:

Hello,

In this case the undergoing chemical reaction is shown on the attached picture whereas cyclohexanol is converted into cyclohexene and water by the dehydrating effect of the sulfuric acid. Thus, for the starting 3 mL of cyclohexanol, the following stoichiometric proportional factor is applied in order to find the theoretical yield of cyclohexene in moles:

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Besides, the mass could be computed as well by using the molar mass of cyclohexene:

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Even thought, the volume could be also computed by using its density:

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Best regards.

4 0

2 years ago

Read 2 more answers