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kari74 [83]
3 years ago
15

What is the mass of the missing reactant in the following reactions, given that the reaction goes to completion? Do not include

units.
- 2.7 g water + ________ g carbon dioxide --> 9.3 g carbonic acid
- 32.0 g sodium hydroxide + 16.0 g hydrofluoric acid --> ________ g water + 33.6 g sodium fluoride
- 0.60 g calcium carbonate + 0.48 g sodium hydroxide --> ________ g sodium carbonate + 0.45 g calcium hydroxide
- 0.53 g sodium hydroxide + 0.37 g carbon dioxide --> ________ g sodium hydrogen carbonate
Chemistry
1 answer:
N76 [4]3 years ago
6 0
1) <span> 2.7 g water + 6,6 g carbon dioxide </span>→<span> 9.3 g carbonic acid.
According to </span><span>principle of mass conservation mass of reactants and products are the same after chemical reactio. 2,7 g + 6,6 g = 9,3 g.
2) </span><span>32.0 g sodium hydroxide + 16.0 g hydrofluoric acid --> 14,4 g water + 33.6 g sodium fluoride.
m(water) = 32 g + 16 g - 33,6 g.
3) </span><span>0.60 g calcium carbonate + 0.48 g sodium hydroxide --> 0,63 g sodium carbonate + 0.45 g calcium.
m(sodium carbonate) = 0,6 g + 0,48 g - 0,45 g.
4) </span><span>0.53 g sodium hydroxide + 0.37 g carbon dioxide --> 0,9 g sodium hydrogen carbonate.
m(sodium hydrogen carbonate) = 0,53 g + 0,37 g = 0,9 g.</span>
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formic acid buffer containing 0.50 M HCOOH and 0.50 M HCOONa has a pH of 3.77. What will the pH be after 0.010 mol of NaOH has b
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Answer:

pH = 3.95

Explanation:

It is possible to calculate the pH of a buffer using H-H equation.

pH = pka + log₁₀ [HCOONa] / [HCOOH]

If concentration of [HCOONa] = [HCOOH] = 0.50M and pH = 3.77:

3.77 = pka + log₁₀ [0.50] / [0.50]

<em>3.77 = pka</em>

<em />

Knowing pKa, the NaOH reacts with HCOOH, thus:

HCOOH + NaOH → HCOONa + H₂O

That means the NaOH you add reacts with HCOOH producing more HCOONa.

Initial moles of 100.0mL = 0.1000L:

[HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH

[HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa

After the reaction, moles of each species is:

0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH

0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa

With these moles of the buffer, you can calculate pH:

pH = 3.77 + log₁₀ [0.0600] / [0.0400]

<h3>pH = 3.95</h3>

3 0
3 years ago
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