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Viktor [21]
3 years ago
12

Where is the nitrogen cycle Found?

Chemistry
1 answer:
lions [1.4K]3 years ago
8 0
Is found in both the living portion of our planet and the inorganic parts of the earths system
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Why a person bitten by a snake can not drink alcohol​
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Alcoholic liquors are harmful to persons bitten by venomous snakes." The alcohol acts first as a stimulant, speeding up the circulation, quickly distributing the poison through the body. When the effect wears off it becomes a depressant, lowering the victim's resistance,
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3 years ago
Help me pls I put 49 point( every single point that I have) pls help me
Aliun [14]

Answer:

Chemical reaction, compound, molecule, covalent bonds, ionic bonds, ions. (first page)

Oxygen and hydrogen, 1 hydrogen 2 oxygen, H2O, a covalent bond, then mark off is always a liquid. (Second page)

Explanation:

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3 years ago
What is the one variable that is changed in the experimental group called? the placebo the independent variable the control grou
BlackZzzverrR [31]

Answer:

the independant variable

Explanation:

4 0
3 years ago
Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation f
dmitriy555 [2]

<u>Answer:</u> The mass of cryolite produced is 51.48 kg

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For aluminium oxide:</u>

Given mass of aluminium oxide = 12.5 kg = 12500 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of aluminium oxide = 101.96 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium oxide}=\frac{12500g}{101.96g/mol}=122.6mol

  • <u>For NaOH:</u>

Given mass of NaOH = 55.4 kg = 55400 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

\text{Moles of NaOH}=\frac{55400g}{40g/mol}=1389mol

  • <u>For HF:</u>

Given mass of HF = 55.4 kg = 55400 g

Molar mass of HF = 20 g/mol

Putting values in equation 1, we get:

\text{Moles of HF}=\frac{55400g}{20g/mol}=2770mol

For the given chemical reaction:

Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O(g)

By Stoichiometry of the reaction:

1 mole of aluminium oxide reacts with 6 moles of sodium hydroxide and 12 moles of HF.

So, 122.6 moles of aluminium oxide will react with (6\times 122.6)=735.6mol of sodium hydroxide and (12\times 122.6)=1471.2mol of HF

As, given amount of NaOH and HF is more than the required amount. So, they are considered as an excess reagent.

Thus, aluminium oxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aluminium oxide produces 2 moles of cryolite

So, 122.6 moles of aluminium oxide will produce = \frac{2}{1}\times 122.6=245.2mol of cryolite

Now, calculating the mass of cryolite by using equation 1:

Molar mass of cryolite = 209.94 g/mol

Moles of cryolite = 245.2 mol

Putting values in equation 1, we get:

245.2mol=\frac{\text{Mass of cryolite}}{209.94g/mol}\\\\\text{Mass of cryolite}=(245.2mol\times 209.94g/mol)=51477.3g

Converting this into kilograms, we use the conversion factor:

1 kg = 1000 g

So, 51477.3 g\times (\frac{1kg}{1000g})=51.48kg

Hence, the mass of cryolite produced is 51.48 kg

7 0
3 years ago
An oxybromate compound, kbrox, where x is unknown, is analyzed and found to contain 43.66 % br.
aniked [119]

Answer is: an oxybromate compound is KBrO₄ (x = 4).

ω(Br) = 43.66% ÷ 100%.

ω(Br) = 0.4366; mass percentage of bromine.

If we take 100 grams of compound:

m(Br) = ω(Br) · 100 g.

m(Br) = 0.4366 · 100 g.

m(Br) = 43.66 g; mass of bromine.

n(Br) = m(Br) ÷ M(Br).

n(Br) = 43.66 g ÷ 79.9 g/mol,

n(Br) = 0.55 mol; amoun of bromine.

From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).

m(K) = 0.55 mol · 39.1 g/mol.

m(K) = 21.365 g; mass of potassium in the compound.

m(O) = 100 g - 21.365 g - 43.66 g.

m(O) =34.97 g; mass of oxygen.

n(O) = 34.97 g ÷ 16 g/mol.

n(O) = 2.185 mol.

n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.

n(K) : n(Br) : n(O) = 1 : 1 : 4.

6 0
3 years ago
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