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vekshin1
4 years ago
8

A raindrops has a mass of 0.50g. how many moles of water does a raindrop contain

Mathematics
1 answer:
alina1380 [7]4 years ago
7 0
So, here you have to look at the periodic table and see what the molar mass of each molecule composing the water compound H_2 O
its two hydrogen atom plus an oxygen molecule, making it about 18g per mol.

That said, looking at the mass, we have to ask ourselves, if a mol of water contains 18g of mass, how many moles do we have in 0.5g of water? 
We just do a cross multiplication:
\frac{(0.5)}{18} =0.027 mol
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Simplify this expression.<br><br> 20.25 –9 1/2+ 11<br><br> Pls help
grigory [225]

Answer:

The answer is -0.25.

Step-by-step explanation:

Using PEMDAS, add 9.5 + 11 first, which results in 20.5. Then substract 20.5 from 20.25, 20.25 - 20.5. This leads to -0.25 or -1/4.

7 0
3 years ago
Which measure is equivalent to 1.5 kg?
Airida [17]

Answer:

number 2.) 1,500g

Step-by-step explanation:

i just know..

7 0
2 years ago
What is the solution of -5x &gt; 25?
Veseljchak [2.6K]
<span>Divide both sides by -5.
</span><span><span>−<span>5x /</span></span><span>−5  </span></span>>  <span>25 /<span>−<span>5


ANSWER: X < - 5

</span></span></span>
4 0
3 years ago
The capacity of a small theater is 125 people. Tickets for a particular show cost $14.50. The revenue for that show is a functio
faltersainse [42]

Answer:

The domain of this function is first restricted to all non negative integers  because the theater can sell only a whole number of tickets.

The domain is restricted to numbers less than or equal to 125 because the theater cannot sell more tickets than capacity.

Step-by-step explanation:

3 0
3 years ago
- Given a test statistic of t= 3.018 of a left-tailed test with n= 13, use a 0.01 significance level
alina1380 [7]

Using the t-distribution, it is found that the correct option is:

0.005 < P-value < 0.01, reject the null hypothesis.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if the mean of a given population is equal to 85, that is:

H_0: \mu = 85

We have a left-tailed test, hence, at the alternative hypothesis, it is tested if the mean of the population is less than 85, that is:

H_1: \mu < 85

<h3>What is the decision?</h3>

The test statistic is of t = -3.018.

Using a t-distribution calculator, with test statistic t = -3.018, 13 - 1 = 12 df and a significance level of 0.01, it is found that the p-value for a left-tailed test is of 0.0051.

The p-value is <u>less than 0.01</u>, hence the null hypothesis is rejected and the correct option is:

0.005 < P-value < 0.01, reject the null hypothesis.

More can be learned about the t-distribution at brainly.com/question/16313918

5 0
2 years ago
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