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Ede4ka
3 months ago
7

What is the empirical formula of an oxide of chromium that is 48 percent oxygen by mass?

Mathematics
1 answer:
Svet_ta [12.7K]3 months ago
4 0

Answer:

CrO₃.

Step-by-step explanation:

First, we will find the chromium percentage in the oxide by using the following process:

Oxygen (O) = 48%

Chromium (Cr) =?

The oxide consists solely of chromium and oxygen, and its chromium percentage is calculated as:

Cr = 100 – percentage of oxygen

Cr = 100 – 48

Cr = 52%

Next, we will determine the empirical formula for the oxide:

Chromium (Cr) = 52%

Oxygen (O) = 48%

Now, we divide by their molar mass:

Cr = 52/52 = 1

O = 48/16 = 3

Now, divide by the lowest value:

Cr = 1/1 = 1

O = 3/1 = 3

Thus, the empirical formula for the oxide is CrO₃.

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Response with clarification:

Let p denote the proportion of adults in the town who have encountered this flu strain.

According to the provided information

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∵ H_a this is a two-tailed test.

Test statistic:

z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}

, where p= denotes the population proportion

{\hat{p} = signifies the sample proportion

n= represents the sample size

Setting n= 6 and {\hat{p}=\dfrac{6}{88}\approx0.068 and p=0.08

z=\dfrac{0.068-0.08}{\sqrt{\dfrac{0.08(1-0.08)}{88}}}

z=\dfrac{-0.012}{0.0289199522192}\approx-0.415

P-value for the two-tailed test:[2P(Z>|z|)

=2P(Z>|-0.415|)

=2P(Z>0.415) = 2[1-P(Z≤0.415)] [∵ P(Z>z)=1-P(Z≤z)]

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This leads us to conclude that there is insufficient evidence to back the assertion that the percentage of all adults in this town exposed to this flu strain deviates from the national average of 8%.

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The null and alternative hypotheses are:

H_0: \mu=15\\\\H_a: \mu\neq15

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The degrees of freedom are calculated as df=(20-1)=19.

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