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Nataly [62]
3 years ago
13

What is carbon monoxide

Chemistry
2 answers:
saw5 [17]3 years ago
4 0

Answer:

hope this helps

Explanation:

Carbon monoxide (CO) is an odorless, colorless gas formed by the incomplete combustion of fuels. When people are exposed to CO gas, the CO molecules will displace the oxygen in their bodies and lead to poisoning.

marishachu [46]3 years ago
4 0

Answer:

Carbon monoxide (chemical formula CO) is a colorless, odorless, tasteless, flammable gas that is slightly less dense than air. Carbon monoxide consists of one carbon atom and one oxygen atom. It is the simplest molecule of the oxocarbon family. In coordination complexes the carbon monoxide ligand is called carbonyl.

Explanation:

.from Wikipedia

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What is the electron configuration of Iron
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The electron configuration of Iron is:

[Ar] 3d⁶4s²

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A body-centered cubic unit cell has an edge length of 588 pm. What is the radius of the atoms composing the unit cell?
blagie [28]

Answer:

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3 years ago
Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the eq
ad-work [718]

Answer:

-75 cm^3/min

Explanation:

Given from Boyle's law;

PV=C

From product rule;

VdP/dt + PdV/dt = dC/dt

but dC/dt = 0, V= 500 cm^3, P= 200kPa, dP= 30kPa/min

PdV/dt = dC/dt - VdP/dt

dV/dt = dC/dt - VdP/dt/ P

substituting values;

dV/dt = 0 - (500 * 30)/200

dV/dt = -75 cm^3/min

5 0
2 years ago
The products in a decomposition reaction _____.
alex41 [277]
I believe the answer is  <span>can be elements or compounds
In this case, elements in the decomposition reaction is the substance that cannot be separated into simpler substances.
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8 0
2 years ago
Read 2 more answers
Heyy guys, so basically i need help with stoichiometric calculation I will give you 100 points just to answer all of these answe
jeka94

Answer:

3. The mass of ethanol required is approximately 0.522869 g

The mass of ethanoic acid required is approximately 0.68156 g

4. The mass of iron (III) oxide required is approximately 285.952.189.095 tonnes

5. The mass of silver nitrate required is approximately 14.53 grams

6. The mass of copper oxide that would be needed is approximately 31.86 grams

7. a. The mass of the precipitate, Zn(OH)₂ formed is approximately 49.712 grams

b. The mass of the precipitate, Al(OH)₃ formed is approximately 13 grams

c. The mass of the precipitate, Mg(OH)₂, formed is approximately 14.579925 grams

Explanation:

3. The 1 mole of ethanol and 1 mole of ethanoic acid combines to form 1 mole of ethyl ethanoate

The number of moles of ethyl ethanoate in 1 gram of ethyl ethanoate, n = 1 g/(88.11 g/mol) = 1/88.11 moles

∴ The number of moles of ethanol = 1/88.11 moles

The number of moles of ethanoic acid = 1/88.11 moles

The mass of ethanol = (46.07 g/mol) × 1/88.11 moles = 0.522869 g

The mass of ethanoic acid in the reaction = 60.052 g/mol × 1/88.11 moles ≈ 0.68156 g

4. 1 mole of iron(III) oxide reacts with 1 mole of CO₂ to produce 1 mole of iron

The number of moles in 100 tonnes of iron= 100000000/55.845 = 1790670.60614 moles

The mass of iron (III) oxide required = 159.69 × 1790670.60614 = 285952189.095 g ≈ 285.952.189.095 tonnes

5. The number of moles of NaCl in 5 grams of NaCl = 5 g/58.44 g/mol = 0.0855578371 moles

The mass of silver nitrate required, m = 169.87 g/mol × 0.0855578371 moles ≈ 14.53 grams

6. The number of moles of CuSO₄·5H₂O in 100 g of CuSO₄·5H₂O = 100 g/(249.69 g/mol) ≈ 0.4005 moles

The mass of copper oxide required, m = 79.545 g/mol × 0.4005 moles ≈ 31.86 grams

7. a. The number of moles of NaOH in the reaction = 20 g/(39.997 g/mol) ≈ 0.5 moles

2 moles of NaOH produces 1 mole of Zn(OH)₂

0.5 moles of NaOH will produce 0.5 mole of Zn(OH)₂

The mass of 0.5 mole of Zn(OH)₂ = 0.5 mole × 99.424 g/mol = 49.712 grams

The mass of the precipitate, Zn(OH)₂ formed = 49.712 grams

b. 6 moles of NaOH produces 2 moles Al(OH)₃

20 g, or 0.5 mole of NaOH will produce (1/6) mole of Al(OH)₃

The mass of the precipitate, Al(OH)₃ formed, m = 78 g/mol×(1/6) moles = 13 grams

c. 2 moles of NaOH produces 1 mole of Mg(OH)₂, therefore;

20 g or 0.5 moles of NaOH formed (1/4) mole of Mg(OH)₂

The mass of the precipitate, Mg(OH)₂, formed, m = 58.3197 g/mol × (1/4) moles = 14.579925 grams

3 0
2 years ago
Read 2 more answers
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