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2 years ago

The molar mass of AlCl3 is 133.34 g/mol. How many molecules of AlCl3 are there in 2 g?

1 answer:

5 0

To solve this problem, we have to apply mole-concept to this and we would need the molar mass of AlCl3. 2g of AlCl3 contains 9.034*10^21 molecules.

<h3>Number of Molecules in 2g of AlCl3</h3>

The molar mass of AlCl3 is given as 133.34g/mol

Data;

mass = 2g
molar mass = 133.34g/mol

Using the mole concept, we can equate the molar mass to the number of molecules present in the entity.

$133.34g = 6.023*10^2^3$

If we equate this

$133.34g = 6.023*10^2^3\\2g = x molecules\\x = \frac{2*6.023*10^2^3}{133.34} \\x = 9.034*10^2^1 molecules$

From the calculations above, 2g of AlCl3 contains 9.034*10^21 molecules.

Learn more on mole concept here;

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Find the length and surface area. Width = 2 Height = 12 Volume = 240

blondinia [14]

Answer:

Length: 10

Surface Area: 20

Step-by-step explanation:

Length:

2*12=24

240/24=10

Surface Area:

2*10=20

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3 years ago

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

mojhsa [17]

Answer:

(A) The minimum sample size required achieve the margin of error of 0.04 is 601.

(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.

Step-by-step explanation:

Let us assume that the percentage of support for the candidate, among voters in her district, is 50%.

(A)

The margin of error, MOE = 0.04.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of z for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601$

Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.

(B)

The margin of error, MOE = 0.02.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of z for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401$

Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.

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3 years ago

A simple random sample of size nequals15 is drawn from a population that is normally distributed. The sample mean is found to be

Bond [772]

Answer with explanation:

To test the Significance of the population which is Normally Distributed we will use the following Formula Called Z test

$z=\frac{\Bar X - \mu}{\frac{\sigma}{n}}$

$\Bar X =31.1\\\\ \sigma=6.3\\\\ \mu=25\\\\n=15\\\\z=\frac{31.1-25}{\frac{6.3}{\sqrt{15}}}\\\\z=\frac{6.1\times\sqrt{15}}{6.3}\\\\z=\frac{6.1 \times 3.88}{6.3}\\\\z=3.756$

→p(Probability) Value when ,z=3.756 is equal to= 0.99992=0.9999

⇒Significance Level (α)=0.01

We will do Hypothesis testing to check whether population mean is different from 25 at the alpha equals 0.01 level of significance.

→0.9999 > 0.01

→p value > α

With a z value of 3.75, it is only 3.75% chance that ,mean will be different from 25.

So,we conclude that results are not significant.So,at 0.01 level of significance population mean will not be different from 25.

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3 years ago

M is the midpoint of XY. XM=10w+2 and MY=5w+22 WHAT IS XY??

Harlamova29_29 [7]

Answer:

15w+24

Step-by-step explanation:

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3 years ago

Read 2 more answers

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barxatty [35]

Answer:

250

Step-by-step explanation:

One quarter of 100 would be 25, so adding a zero to 100 (1000) would also mean adding a 0 to 25. Therefore, 250

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2 years ago