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arsen [322]
3 years ago
11

Which of the these is a balanced chemical equation? A) CO(g) + O2(g) → CO2(g) B) CO(g) + 2O2(g) → 2CO2(g) C) 2CO(g) + O2(g) → 2C

O2(g) D) 2CO(g) + 2O2(g) → 2CO2(g)
Chemistry
2 answers:
jenyasd209 [6]3 years ago
6 0
Answer: C.

This is because the atoms on either side of the equation in all the other solutions are not equal. 
Alinara [238K]3 years ago
3 0

hello there the answer is c sir

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Find the mass (in g) of 1.00 atom of lead.
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In fossil comparisons, it helps to know about similar:
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6 0
2 years ago
How many grams of CO₂ can be produced from the combustion of 2.76 moles of butane according to this equation: 2 C₄H₁₀(g) + 13 O₂
Vilka [71]

Answer:

485.76 g of CO₂ can be made by this combustion

Explanation:

Combustion reaction:

2 C₄H₁₀(g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)

If we only have the amount of butane, we assume the oxygen is the excess reagent.

Ratio is 2:8. Let's make a rule of three:

2 moles of butane can produce 8 moles of dioxide

Therefore, 2.76 moles of butane must produce (2.76 . 8)/ 2 = 11.04 moles of CO₂

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5 0
3 years ago
Which solute would be more effective at lowering the freezing point of water: MgCl2 and KNO3? Explain.
Phantasy [73]

Answer:

AlCl₃.

Explanation:

Adding solute to water causes depression of the boiling point.

The depression in freezing point (ΔTf) can be calculated using the relation:

ΔTf = i.Kf.m,

where, ΔTf is the depression in freezing point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kf is the molal depression constant of water.

m is the molality of the solution (m = 1.0 m, for all solutions).

(1) NaCl:

i for NaCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (NaCl) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(2) MgCl₂:

i for MgCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

∴ ΔTb for (MgCl₂) = i.Kb.m = (3)(Kf)(1.0 m) = 3(Kf).

(3) NaCl:

i for KBr = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (KBr) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(4) AlCl₃:

i for AlCl₃ = no. of particles produced when the substance is dissolved/no. of original particle = 4/1 = 4.

∴ ΔTb for (CoCl₃) = i.Kb.m = (4)(Kf)(1.0 m) = 4(Kf).

So, the ionic compound will lower the freezing point the most is: AlCl₃

4 0
3 years ago
Read 2 more answers
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