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miskamm [114]
3 years ago
10

Question

Chemistry
1 answer:
Black_prince [1.1K]3 years ago
7 0

Answer:

efef2w

Explanation:

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4 0
3 years ago
A sample of carbon dioxide at RTP is 0.50 dm3. How many grams of carbon dioxide do we have?
prohojiy [21]

Answer:

0.924 g

Explanation:

The following data were obtained from the question:

Volume of CO2 at RTP = 0.50 dm³

Mass of CO2 =?

Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:

1 mole of gas = 24 dm³ at RTP

Thus,

1 mole of CO2 occupies 24 dm³ at RTP.

Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e

Xmol of CO2 = 0.5 /24

Xmol of CO2 = 0.021 mole

Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.

Finally, we shall determine the mass of CO2 as follow:

Mole of CO2 = 0.021 mole

Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol

Mass of CO2 =?

Mole = mass /Molar mass

0.021 = mass of CO2 /44

Cross multiply

Mass of CO2 = 0.021 × 44

Mass of CO2 = 0.924 g.

3 0
3 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
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