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Archy [21]
2 years ago
10

Dr. Moore has discovered a new vaccine composed of carbon, hydrogen, and oxygen. When the product was purified and sent off for

elemental analysis it gave the following mass percents: 68.85% C, 4.95% H. Determine the empirical formula of this vaccine.
Chemistry
2 answers:
Crazy boy [7]2 years ago
4 0

Answer:

C_7H_6O_2

Explanation:

We have to assume that we have <u>100 g of the unknow</u>  as first step. With this in mind we can <u>calculate the grams of C, H and O</u> in the sample.

100~g~\frac{68.85}{100}=68.85~g~of~C

100~g~\frac{4.95}{100}=4.95~g~of~H

For the calculation of O we have to know the <u>percentage of O</u> therefore we can add the percentages of H and C and then do a <u>substraction from 100 </u>%, so:

%~of~O=~100-(68.85+4.95)=~26.2~%~of~O100-(68.85+4.95)=26.2 %

With this value we can<u> calculate the amount of O in grams</u> in the sample:

100~g~\frac{26.2}{100}=26.2~g~of~O

The next step is the <u>calculation of moles for each atom</u>, for this we have to know the <u>atomic mass</u> of each atom ( C: 12 g/mol; H 1 g/mol; O 16 g/mol):

68.85~g~of~C\frac{1~mol~C}{12~g~C}=~5.74~mol~C

4.95~g~of~H\frac{1~mol~H}{1~g~H}=4.95~mol~of~H

26.2~g~of~O\frac{1~mol~O}{16~g~O}=1.64~mol~of~O

Now we have to <u>divide by the smallest value</u>, so:

\frac{5.74~mol~C}{1.64}=3.5

\frac{4.95~mol~H}{1.64}=3

\frac{1.64~mol~O}{1.64}=1

We obtain an <u>decimal number</u> for C, therefore we have to <u>multiply</u> all by "2", so:

3.5x2= 7~mol~C

3x2= 6~mol~H

1x2= 2~mol~O

Therefore <u>the formula would be</u>: C_7H_6O_2

pantera1 [17]2 years ago
3 0

Answer:

C4H3O

Explanation:

Data obtained from the question include:

Carbon (C) = 68.85%

Hydrogen (H) = 4.95%

Since the vaccine contains carbon, hydrogen and oxygen, the remaining percentage will be for oxygen. The percentage of oxygen will be:

Oxygen (O) = 100 - (68.85 + 4.95)

= 26.2%

The empirical formula for the vaccine can be obtained as follow:

Carbon (C) = 68.85%

Hydrogen (H) = 4.95%

Oxygen (O) = 26.2%

Divide the above by their individual molar mass as shown below:

C = 68.85/12 = 5.7375

H = 4.95/1 = 4.95

O = 26.2/16 = 1.6375

Next, divide by the smallest number as shown below:

C = 5.7375/1.6375 = 4

H = 4.95/1.6375 = 3

O = 1.6375/1.6375 = 1

Therefore, the empirical formula for the vaccine is C4H3O

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Answer:- The hydroxide ion concentration of the solution is 1.95*10^-^6 .

Solution:- The formula used to calculate pOH from hydroxide ion is:

pOH=-log[OH^-]

When pOH is given and we are asked to calculate hydroxide ion concentration then we multiply both sides by negative sign and take antilog and what we get on doing this is:

[OH^-]=10^-^p^O^H

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[OH^-]=10^-^5^.^7^1

[OH^-] = 0.00000195 or 1.95*10^-^6

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3 0
3 years ago
HELP
Jobisdone [24]

Answer:

hello,

first one is 25.15

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Explanation:

honestly if you look up on the internet there is a converter to get you your answers

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Let's consider the following reaction.

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a.

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We can find the average rate of the reaction over this time interval using the following expression.

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Greeley [361]

Answer:

A chemical change occurred, a gas and precipitate was produced

Explanation:

From the question , we are informed of science lab, where Cash mixes two clear liquids together in a beaker. Bubbles are produced, and a white solid forms and settles to the bottom.

In this case the change that took place is chemical change ( is one where new product are formed after two substance react) the bubbles that is produced signify the presence of gas in the product, white solid formed is reffered to as a precipitate( which is reffered to as solid that is been formed from a particular solution).

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