Which property describes a material's ability to conduct heat?

1. A gas having the following composition is burnt under a boiler with 50% excess air.

jeka94

The composition of the stack gas are :

$CH_4$ = 0.8713

$C_3H_8$ = 0.0202

CO = 0.107

<h3 /><h3>What is a mole fraction?</h3>

The ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all of the components.

Assuming 100 g of the stack gas. Calculate the mass of each species in this sample according to their percentages.

Mass of $CH_4$ : 70% of 100 g = 70 g

Mass of $C_3H_8$ : 15% of 100 g = 15 g

Mass of CO : 15% of 100 g = 15 g

Now calculate the number of moles of each species:

Number of moles of $CH_4$ : $\frac{70 g}{16.04 g/mol}$ = 4.3 mole

Number of moles of $C_3H_8$ : $\frac{15 g}{144.1 g/mol}$ = 0.10 mole

Mass of CO : $\frac{15 g}{28.01 g/mol}$ = 0.53 mole

Now to calculate the mole fraction of each we use the formula:

Mole fraction of $CH_4$ : $\frac{4.3}{4.935}$ = 0.8713

Mole fraction of $C_3H_8$ : $\frac{0.10}{4.935}$ = 0.0202

Mole fraction of CO : $\frac{0.53}{4.935}$ = 0.107

Hence, composition of the stack gas are:

$CH_4$ = 0.8713

$C_3H_8$ = 0.0202

CO = 0.107

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8 0

3 years ago

The Environmental Protection Agency has determined that safe drinking

dmitriy555 [2]

Answer:

b) $\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$

The confidence interval for this case is given (6.21, 6.59)

So we can conclude at 95% of confidence that the true mean for the PH concentration is between 6.21 and 6.59 moles per liter

c) Since the confidence interval not contains the value 7 we reject the hypothesis that the true mean is equal to 7. And the same result was obtained with the t test for the true mean.

Explanation:

We assume that part a is test the claim. And we can conduct the following hypothesis test:

Null hypothesis: $\mu =7$

Alternative hypothesis $\mu \neq 7$

The statistic is to check this hypothesi is given by:

$t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}$

We know the following info from the problem:

$\bar X = 6.4 , s=0.5, n =30$

Replacing we got:

$t = \frac{6.4-7}{\frac{0.5}{\sqrt{30}}}= -6.573$

And the p value would be:

$p_v= 2*P(Z$

Since the p value is very low compared to the significance assumed of 0.05 we have enough evidence to reject the null hypothesis that the true mean is equal to 7 moles/liter

Part b

The confidence interval is given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$

The confidence interval for this case is given (6.21, 6.59)

So we can conclude at 95% of confidence that the true mean for the PH concentration is between 6.21 and 6.59 moles per liter

Part c

Since the confidence interval not contains the value 7 we reject the hypothesis that the true mean is equal to 7. And the same result was obtained with the t test for the true mean.

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