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iogann1982 [59]
2 years ago
8

If a typical antacid tablet contains 2.0 g of sodium hydrogen carbonate, how many moles of carbon dioxide should one tablet yiel

d?
Chemistry
2 answers:
Sphinxa [80]2 years ago
7 0

Answer: The 2.0 grams of sodium hydrogen carbonate will yield 0.0238 moles of carbon-dioxide.

Explanation:

Mass of sodium hydrogen carbonate in 1 tablet = 2.0 g

NaHCO_3+ HCl\rightarrow NaCl+H_2O+CO_2

Moles of NaHCO_3=\frac{\text{Mass of }Na_2CO_3}{\text{Molar Mass of }Na_2CO_3}=\frac{2.0 g}{84 g/mol}=0.0238 moles

According to reaction 1 mole ofNa_2CO_3 gives 1 mole of CO_2 then 0.0238 mole ofNa_2CO_3 will give: \frac{1}{1}\times 0.0238 moles of CO_2 that is 0.0238 moles.

So, the 2.0 grams of sodium hydrogen carbonate will yield 0.0238 moles of carbon-dioxide.

ivolga24 [154]2 years ago
6 0
The equation of the chemical reaction is NaHCO3 + H+ --> H2O + CO2 + Na
To determine the total number of moles of carbon dioxide, the given mass of sodium hydrogen carbonate is divided with its own molar mass. Then it is multiplied with the ratio between NaHCO3 and carbon dioxide. The total number of moles of CO2 one tablet should yield is 0.024 mole.
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Recall that your hypothesis is that these values are the fraction of atoms that are still radioactive after n half-life cycles.
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Answer : A= 0.5, B = 0.25 , C = 0.125, D = 0.015625 and E = 0.00390625

Explanation :

Half life of a substance is defined as the amount of time taken by the substance to reduce to half of its original amount.

Here n represents the number of half lives.

The amount of substance that remains after n half lives can be calculated using the given formula, 0.5^{n}

So when we have n =1,

Fraction of substance that remains = 0.5¹ = 0.5.

That means after first half life over, the amount of substance that remains is 0.5 times that of original.

Therefore we have A = 0.5

When n = 2, we have 0.5² = 0.25

So when 2 half lives are over, the amount of substance that remains is 0.25 times that of original

Therefore B = 0.25

When n = 3, we have 0.5³ = 0.125

So when 3 half lives are over, the amount of substance that remains is 0.125 times that of original.

Therefore we have C = 0.125

When n = 6 , we have 0.5⁶ = 0.015625

So D = 0.015625

When n = 8, we have 0.5⁸ = 0.00390625

Therefore E = 0.00390625

The values for A, B, C, D and E are 0.5, 0.25, 0.125, 0.015625 and 0.00390625 respectively.

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siniylev [52]

Answer:

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What role do the coefficients serve in a chemical reaction?
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Determine whether or not the mixing of each of the two solutions indicated below will result in a buffer.
WARRIOR [948]
Part A

75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF

This combination will form a buffer.

Explanation

Here, weak acid HF and its conjugate base F- is available in the solution

Part B

150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl

This combination cannot form a buffer.

Explanation

Here, moles of HF = 0.15 x 0.1 = 0.015 moles

Moles of HCl = 0.135 x 0.175 = 0.023

Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution

Part C

165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH

This combination will form a buffer.

Explanation

Moles of HF = 0.165 x 0.1 = 0.0165 moles

Moles of KOH = 0.135 x 0.05 = 0.00675 moles

Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer

Part D

125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl

This combination will form a buffer

Explanation

Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer

Part E

105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl

This combination will form a buffer

Explanation

Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles

Moles of HCl = 0.095 x 0.1 = 0.0095 moles

Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer
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2 years ago
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