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sergiy2304 [10]
2 years ago
11

How many moles in 30.0 grams of h3po4

Chemistry
1 answer:
jok3333 [9.3K]2 years ago
5 0
Molar mass H₃PO₄ = 98.0 g/mol

1 mole ----- 98.0 g
? mole ------ 30.0 g

moles = 30.0 * 1 / 98.0

= 0.306 moles

hope this helps!


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Please help!!! ASAP
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1*23=23. 2*12=24. 6*16=96

23+24+96=143

(23*100)/143. (24*100)/143. (96*100)/143

=16%. =16.7%. =67.1%

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What is a chemical off of the periodic table that is sticky
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I think it’s the one that has a Br
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Which formula correctly represents the product of an addition reaction between ethene and chlorine?
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The formula that correctly represents the product of an addition reaction between ethene and chlorine is  C2H4Cl2
Addition reaction occurs when an atom is added to a compound that has a double bond or triple bond (unsaturated hydrocarbons). Unsaturated compounds are associated with addition reactions. For example Ethene is an example of unsaturated hydrocarbon; when reacted with chlorine gas , chlorine atoms are added to each carbon atoms. 
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What will you expect from an endothermic reaction
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An endothermic reaction is when the energy is absorbed, while an exothermic reaction releases energy.
3 0
3 years ago
A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of
Stella [2.4K]

Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH}  =0.1900gO

4. Compute the moles of oxygen by using its molar mass:

n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1

6. Search for the closest whole number (in this case multiply by 2):

C_3H_6O_2

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

M=12*3+1*6+16*2=74g/mol

Which is about three times in the molecular formula, for that reason, the actual formula is:

C_9H_{18}O_6

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

6 0
3 years ago
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