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Brilliant_brown [7]
3 years ago
9

What would happen to the following endothermic reaction that is in equilibrium if heat is added? N2O4 (g) Two arrows stacked on

top of each other. The top arrow points to the right. The bottom arrow points to the left. 2 NO2 (g):
1) The reaction would not be affected

2) The equilibrium would shift to the left

3) The equilibrium would shift to the right

4) The product concentration would equal the reactants
Chemistry
2 answers:
ludmilkaskok [199]3 years ago
7 0
<span>The event that will happen to the following endothermic reaction that is in equilibrium if heat is added is that </span><span>The equilibrium would shift to the right. The correct answer from your choices is number 3.</span>
Ber [7]3 years ago
6 0
The reaction N2O4 (g) <--> 2NO2 (g) is endothermic, meaning that it consumes heat to move towards formation of the products.
According to Le Chatelier's Principle, therefore, if heat is added, more product (NO2) will be produced, and equilibrium would shift towards the right side. This is choice 3.
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The songs of birds are part of a ritual to attractpotential partners of their own species. The song presents specific patterns r
liberstina [14]

Answer:

Explanation:

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6 0
3 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
Metals react with ______ to form compounds that are alkaline.
kondor19780726 [428]

Metals react with ______ to form compounds that are alkaline.

A. metalloids

B. oxygen (O)

C. non-metals

D. hydrogen (H)

The answer is D, Hydrogen (H).

7 0
3 years ago
Read 2 more answers
A hydrocarbon contains 85.7% carbon and the remainder
Viefleur [7K]

Answer:

CH₂ ;  67.1 %

Explanation:

To determine the empirical formula we need to find what the mole ratio is in whole numbers of the atoms in the compound. To do that we will first need the atomic weights of C and H and then perform our calculation

Assume 100 grams of the compound.

# mol C = 85.7 g / 12.01 g/mol = 7.14 mol

# mol H = 14.3 g /  1.008 g/mol = 14.19 mol

The proportion is 14.9 mol H/ 7.14 mol C = 2 mol H/ 1 mol C

So the empirical formula is CH₂

For the second part we will need to first calculate the theoretical yield for the 12.03 g NaBH₄  reacted and then calculate the percent yield given the 0.295 g B₂H₆ produced.

We need to calculate the moles of  NaBH₄ ( M.W = 37.83 g/mol )

1.203 g  NaBH₄ / 37.83 g/mol =  0.0318 mol

Theoretical yield from balanced chemical equation:

0.0318 mol NaBH₄ x 1 mol B₂H₆ / mol NaBH₄ = 0.0159 mol B₂H₆

Theoretical mass yield B₂H₆ = 0.0159 mol x 27.66 g/ mol =  0.440 g

% yield = 0.295 g/ 0.440 g x 100 = 67.1 %

6 0
3 years ago
The orbit closest to the nucleus has ___________ energy.
deff fn [24]

Answer:

inonic bonds with cavalent bonds

Explanation:

ionic bonds

5 0
3 years ago
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