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IRINA_888 [86]
4 years ago
7

Assume that consumer spending increased by $4 billion while unsold business production aimed at consumers increased by $6 billio

n last year. As a result of these changes, GDP would A. increase by $2 billion. B. increase by $4 billion. C. increase by $10 billion. D. increase by $6 billion. E. decrease by $2 billion.
Chemistry
1 answer:
Natali5045456 [20]4 years ago
5 0

Answer:

The correct option is;

B. Increase by $4 billion

Explanation:

In Gross Domestic Product, (GDP) calculation based on the spending or expenditure approach, the spending components of the countries different economic groups are calculated

The formula for the GDP can be expressed as follows;

GDP = C + G + I + NX

Where;

C = Consumer spending or private sector consumption

G = Spending by the government

I = Economic investment

NX = Net export of goods by the economy.

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A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
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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.

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