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Viefleur [7K]
2 years ago
3

If the rate constant for the forward reaction in N2O4(g)Colorless⇋2NO2(g)Brown is larger than the rate constant for the reverse

reaction, will the constant in [NO2]2[N2O4]=kfkr be greater than 1 or smaller than 1?
Chemistry
1 answer:
igomit [66]2 years ago
7 0

Answer:

\dfrac{k_{f}}{k_{r}} > 1

Explanation:

  N₂O₄(g) ⇋ 2NO₂(g)

colourless     brown

\rm N_{2}O_{4}\xrightarrow[k_{\text{r}}]{k_{\text{f}}}2NO\\K_{\text{eq }} = \dfrac{k_{f}}{k_{r}}\\\\\text{If }k_{f} > k_{r}, K_{\text{eq }} > 1

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2 years ago
A 0.1064 g sample of a pesticide was decomposed by the action of sodium biphenyl. The liberated Cl- was extracted with water and
Ilya [14]

Answer:

Percentage of an aldrin in the sample is 44.41%.

Explanation:

Cl^-+AgNO_3\rightarrow AgCl+NO_{3}^-

Molarity of the silver nitrate solution = 0.03337 M

Volume of the silver nitrate = 23.28 mL = 0.02328 L

Moles of silver nitrate = n

0.03337 M=\frac{n}{0.02328 L}

n = 0.03337 M\times 0.02328 L=0.0007768 mol

According to reaction 1 mole of silver nitrate recats with 1 moles of chloride ions.

Then 0.0007768 moles of silver nitrate will react with:

\frac{1}{1}\times 0.0007768 mol=0.0007768 mol chloride ions.

In one mole of aldrin there are 6  moles of chloride ions.

Then moles of aldrin containing 0.0007768 moles chloride ions are:

\frac{0.0007768 mol}{6}=0.0001295 mol

Moles of aldrin present in the sample = 0.0001295 mol

Mass of 0.0001295 moles of aldrin present in the sample :

0.0001295 mol × 364.92 g/mol =0.04726 g

Percentage of an aldrin in the sample:

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8 0
2 years ago
29. A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C12H22O11) per serving size of 60.0 grams. How many servi
adoni [48]

Answer:

The answer is: 51.8 g (86% of serving size)

Explanation:

In order to solve the problem, we have to first determine the number of moles there are in 11.0 g of sucrose. Sucrose has a molecular weight of 342 g (we calculate this from the molar mass of the elements : 12 x 12 g/mol C + 22 x 1 g/mol H + 11 x 16 g/mol O). So, we divide the mass (11.0 g) into the molecular weight of sucrose:

11.0 g sucrose x 1 mol/342 g sucrose= 0.032 mol

We have 0.032 mol of sucrose in a serving of 60 g. But we need less moles (0.0278 mol):

0.032 mol ------------ 60 g serving

0.0278 mol------------ x= 0.0278 mol x 60 g serving/0.032 mol

                                x= 51.8 g

So,  lesser than 1 serving of 60 g must be eaten to consume 0.0278 mol os sucrose. Exactly, 51.8 g (which stands for a 86% of the serving size).

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2 years ago
A 6.0M solution HCl is diluted to 1.0M How many milliliters of the 6.0M solution would be used to prepare 100.o mL of the dilute
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