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ELEN [110]
3 years ago
5

Under the gold standard, a country in balance-of-trade equilibrium earns income from exports that is equal to the money its resi

dents pay for imports. Group of answer choices True False
Chemistry
1 answer:
solong [7]3 years ago
4 0

The answer is true.

Explanation:

The balance of trade is nothing but the country's exports minus the country's imports.

Exports means, what you produce in the country and sell it to other countries, whereas imports means what you get or buy from the other countries.

When you export more than you import, you have trade surplus .In that case the income from exports are more than the money spent. So you have a trade surplus.

When you import then you have a trade deficit or your income is low. Most of the countries want a trade surplus.

But when the Income from exports and the money spent on imports are the same , the situation is that of balance of trade equilibrium, where the income from exports is equal to the money its residents pay for the imports.

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nalin [4]

Answer:

B

Explanation:

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3 years ago
To begin, follow these steps:
s344n2d4d5 [400]

Answer: 6.2  grams of the sodium acetate can dissolve in 5 milliliters of water. if 124 grams of the sodium acetate dissolves in 100 milliliters of water, then 6.2 grams of the sodium acetate can dissolve in 5 milliliters of water.

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3 years ago
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Calculate the work, w, gained or lost by the system when a gas expands from 15 L to 45 L against a constant external pressure of
marta [7]

Answer:

Work done by the system = 4545 J

Explanation:

The expression for the calculation of work done is shown below as:

w=-P\times \Delta V

Where, P is the pressure

\Delta V is the change in volume

From the question,  

\Delta V = 45 - 15 L = 30 L

P = 1.5 atm

w=-1.5\times 30\ atmL

Also, 1 atmL = 101 J

So,  

w=-1.0\times0.75\times 101\ J=-4545\ J (negative sign implies work is done by the system)

<u>Work done by the system = 4545 J</u>

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3 years ago
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3 years ago
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In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several o
Kitty [74]

Answer:

The estimated feed rate of logs is 14.3 logs/min.

Explanation:

The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.

That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.

Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.

To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.

V=V_c+V_w\\\\M/\rho=M_c/\rho_c+Mw/\rho_w\\\\M/\rho=0.55*M/\rho_c+0.45*M/\rho_w\\\\1/\rho=0.55/\rho_c +0.45/\rho_w\\\\0.55/\rho_c=1/\rho-0.45/\rho_w\\\\0.55/\rho_c=1/(0.64*\rho_w)-0.45/\rho_w=(1/\rho_w)*(\frac{1}{0.64}-\frac{0.45}{1}  )\\\\0.55/\rho_c=1.1125/\rho_w\\\\\rho_c=\frac{0.55}{1.1125}*\rho_w= 0.494*\rho_w

The specific gravity of the wood chips is 0.494.

The average volume of a log is

V_l=(\pi*D^{2} /4)*L=(3.1416*\frac{8^{2}  \, in^{2} }{4} )*9ft*(\frac{12 in}{1ft})= 21714 in^{3}=12.57 ft^{3}

The weight of one log is

M=\rho*V=0.494*\rho_w*12.57  ft^{3}\\\\M=0.494*62.4\frac{lbm}{ft^{3} }*12.57ft^{3}\\\\M=387.5lbm

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n=\frac{supply}{M_{log}}=\frac{3617 tons/day}{387.5 lbm}*\frac{2204lbm}{1ton}\\\\n=20573 logs/day=14.3 logs/min

The feed rate of logs is 14.3 logs/min.

7 0
3 years ago
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