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slamgirl [31]
3 years ago
6

How much energy is required to vaporize 185 g of butane at its boiling point? The heat of vaporization for butane is 23.1 kJ/mol

. Express your answer to three significant figures and include the appropriate units.
Mathematics
1 answer:
arlik [135]3 years ago
6 0

Answer:

73.5 kJ

Step-by-step explanation:

Mass of butane = 185 g

Heat of vaporization for butane = 23.1 kJ/mol

Molar mass of butane = 58.12 g/mol

Number of moles of butane = \frac{\text{Mass of butane}}{\text{Molar mass of butane}}=\frac{185}{58.12}=3.18\ moles

Energy required for burning 185 g of butane = 3.18×23.1 = 73.5 kJ

∴ Energy is required to vaporize 185 g of butane at its boiling point is 73.5 kJ

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A clinical psychologist wants to test whether experiencing childhood trauma reduces one's self-efficacy in adulthood. He randoml
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No, there is not any sufficient evidence to conclude that individuals who have experienced childhood trauma have lower self-efficacy in adulthood.

Step-by-step explanation:

We are given that a clinical psychologist wants to test whether experiencing childhood trauma reduces one's self-efficacy in adulthood.

He randomly selects 22 adults who have experienced childhood trauma and finds that their mean self-efficacy score equals 118.1.

Self-efficacy scores in the general population of adults are distributed normally with a mean equal to 118.5 and a standard deviation equal to 18.8 .

<em>Let </em>\mu<em> = mean self-efficacy score.</em>

So, Null Hypothesis, H_0 : \mu \geq 118.5      {means that the individuals who have experienced childhood trauma have higher or same self-efficacy in adulthood}

Alternate Hypothesis, H_A : \mu < 118.5    {means that the individuals who have experienced childhood trauma have lower self-efficacy in adulthood}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;

                      T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean self-efficacy score = 118.1

            \sigma = population standard deviation = 18.8

          n = sample of adults who have experienced childhood trauma = 22

So, <u><em>test statistics</em></u>  =  \frac{118.1-118.5}{\frac{18.8}{\sqrt{22} } }  

                              =  -0.0998

The value of z test statistics is -0.0998.

Since, in the question we are not given with the level of significance so we assume it to be 5%. <u>Now, at 5% significance level the z table gives critical value of -1.645 for left-tailed test.</u>

Since our test statistic is more than the critical value of z as -0.0998 > -1.645, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the individuals who have experienced childhood trauma have higher or same self-efficacy in adulthood.

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