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Nana76 [90]
3 years ago
10

Given that sodium chloride is 39.0% sodium by mass, how many grams of sodium chloride are needed to have 100.mg of Na present? G

iven that sodium chloride is 39.0% sodium by mass, how many grams of sodium chloride are needed to have of Na present? a. 0.256 b. 256 c. 39.0 d. 0.0390 e. none of the above
Chemistry
1 answer:
seraphim [82]3 years ago
8 0

<u>Answer:</u> The mass of sodium chloride solution present is 0.256 grams.

<u>Explanation:</u>

We are given:

39.0 % of sodium in sodium chloride solution

This means that 39.0 grams of sodium is present in 100 grams of sodium chloride solution

Mass of sodium given = 100 mg = 0.1 g     (Conversion factor:  1 g = 1000 mg)

Applying unitary method:

If 39 grams of sodium metal is present in 100 grams of sodium chloride solution

So, if 0.1 grams of sodium metal will be present in = \frac{100}{39}\times 0.1=0.256g of sodium chloride solution.

Hence, the mass of sodium chloride solution present is 0.256 grams.

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Which of the following is a salt that will form from the combination of a strong base with a weak acid?
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Answer:

A. NaHCO₃

Explanation:

NaHCO₃ ⇒ NaOH + H₂CO₃

NaOH is a strong base and H₂CO₃ is a weak acid. Therefore, NaHCO₃ is a salt of a strong base-weak acid reaction. The salt is basic because carbonic acid (H₂CO₃) is a weak acid so it remains undissociated. So, there is a presence of additional OH⁻ ions that makes the solution basic.

Hope that helps.

8 0
3 years ago
A chemist needs 0.550 mol selenium for a reaction. What mass of selenium should a chemist use?
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Answer:

The chemist would require to use 43.43 grams.

Explanation:

In order to solve this problem we need to know<u> how much do 0.550 moles of selenium weigh</u>. To do that we use selenium's<em> molar mass </em>and multiply it by the given number of moles:

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4 0
2 years ago
A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa is:

pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

4 0
3 years ago
HELP ASAP ILL GIVE BRAINLIEST!!
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Of course we would experience them.

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