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yawa3891 [41]
3 years ago
10

The yearly amounts of carbon emissions from cars in Belgium are normally distributed with a mean of 13.9 gigagrams per year and

a standard deviation of 5.8 gigagrams per year. Find the probability that the amount of carbon emissions from cars in Belgium for a randomly selected year are between 11.5 gigagrams and 14.0 gigagrams per year. a. 0.340 b. 0.660 c. 0.167 d. 0.397
Chemistry
1 answer:
zavuch27 [327]3 years ago
4 0

Answer:

The correct option is;

c. 0.167

Explanation:

The parameters given are;

The mean, μ = 13.9 Gigagrams/year

The standard deviation, σ = 5.8 Gigagrams/year

The z-score formula is given as follows;

z = \dfrac{x - \mu }{\sigma }

Where:

x = Observed score = 11.5 Gigagrams/year

We have;

z = \dfrac{11.5 - 13.9 }{5.8 } =  \dfrac{-2.4 }{5.8 } = 0.4138

From the z-score table relations/computation, the probability (p-value) = 0.6605

Where:

x = Observed score = 14 Gigagrams/year

We have;

z = \dfrac{14- 13.9 }{5.8 } =  \dfrac{0.1}{5.8 } = 0.01724

From the z-score table relations/computation, the probability (p-value) = 0.4931

Therefore, the probability, p_{ca}, that the amount of carbon emissions from cars in Belgium for a randomly selected year are between 15.5 Gigagrams/year and 14 Gigagrams/year = The area under the normal curve bounded by the p-values for the two amounts of carbon emission

Which gives;

p_{ca} = 0.6605 - 0.4931 = 0.1674 ≈ 0.167

Therefore, the correct option is c. 0.167.

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Answer: 1. 2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu

2. 3 moles of CuCl_2 : 2 moles of Al

3. 0.33 moles of CuCl_2 : 0.92 moles of Al

4. CuCl_2 is the limiting reagent and Al is the excess reagent.

5. Theoretical yield of AlCl_3 is 29.3 g

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Al=\frac{25.0g}{27g/mol}=0.92moles

\text{Moles of} CuCl_2=\frac{45.0g}{134g/mol}=0.33moles

The balanced chemical equation is:

2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu  

According to stoichiometry :

3 moles of CuCl_2 require = 2 moles of Al

Thus 0.33 moles of CuCl_2 will require=\frac{2}{3}\times 0.33=0.22moles  of Al

Thus CuCl_2 is the limiting reagent as it limits the formation of product and Al is the excess reagent.

As 3 moles of CuCl_2 give = 2 moles of AlCl_3

Thus 0.33 moles of CuCl_2 give =\frac{2}{3}\times 0.33=0.22moles  of AlCl_3

Theoretical yield of AlCl_3=moles\times {\text {Molar mass}}=0.22moles\times 133.34g/mol=29.3

Thus 29.3 g of aluminium chloride is formed.

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2 years ago
Which direction will the following reaction (in a 5.0 L flask) proceed if the pressure of CO_2(g) is 1.0 atm? CaCO_3(s) rightarr
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Answer:

d. To the left because Q > K_p

Explanation:

Hello,

In this case, for the given reaction:

CaCO_3(s) \rightarrow CaO(s) + CO_2(g)

The pressure-based equilibrium expression is:

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In such a way, since Kp is given we rather compute the reaction quotient at the specificed pressure of carbon dioxide as shown below:

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Therefore, since Q>Kp we can see that there are more products than reactants, which means that the reaction must shift leftwards towards the reactants in order to reestablish equilibrium, thus, answer is d. To the left because Q > Kp.

Regards.

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