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3 years ago

In some developing nations, the percentage of animal protein from fish in the diet can be as high as _____.

Chemistry

1 answer:

Makovka662 [10]3 years ago

5 0

Answer:

20%

Explanation:

The fish consumption has had lot of variations in the past few centuries, most noticeably in the last century. Initially it was the richer countries that had population which was consuming the most animal protein from fish in the world, while the developing nations had less. Things have changed though, especially during the 1990's, when the developing nations experienced a rapid increase in the consumption of fish. This was largely due to the growing standard and the availability of fish, so in some of the developing nations the percentage of animal protein from fish reached around 20%.

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How many grams of nh3 can be produced from 3.72 mol of n2 and excess h2.

olganol [36]

N₂ + 3H₂ ⇒ 2NH₃
1mol : 2mol
3,72mol : 7,44mol

n = 7,44mol
M = 17g/mol

m = n * M = 7,44mol * 17g/mol = 126,48g

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3 years ago

How to convert 250 in scientific notation

cricket20 [7]

In scientific notation, a number is less than ten but more than one.

Move the decimal point from 0, 250.000 <- this is the same as 250 to between 2 and 5.

I had to move two spaces.

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I hope this helps!
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3 years ago

If you feed 100 kg of N2 gas and 100 kg of H2 gas into a

torisob [31]

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of $N_2$ = 100 kg = 100000 g

Mass of $H_2$ = 100 kg = 100000 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $H_2$ = 2 g/mole

Molar mass of $NH_3$ = 17 g/mole

First we have to calculate the moles of $N_2$ and $H_2$ .

$\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles$

$\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2+3H_2\rightarrow 2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $N_2$ react with 3 mole of $H_2$

So, 3571.43 moles of $N_2$ react with $3571.43\times 3=10714.29$ moles of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $N_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NH_3$

From the reaction, we conclude that

As, 1 mole of $N_2$ react to give 2 mole of $NH_3$

So, 3571.43 moles of $N_2$ react to give $3571.43\times 2=7142.86$ moles of $NH_3$

Now we have to calculate the mass of $NH_3$

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

$\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg$

Therefore, the mass of ammonia produced can be, 121.429 kg

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3 years ago

Sodium nitrate and lead (ii) acetate express your answer as a chemical equation. identify all of the phases in your answer. ente

ICE Princess25 [194]

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The compounds which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

Thus the exchange of ions take place and all the compounds are soluble so the chemical formulas are followed by the symbol (aq).

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spayn [35]

Answer:

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Explanation:

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