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Dmitry_Shevchenko [17]
3 years ago
9

What substances are in H2CO3(aq) H2O(I) + CO2(g)

Chemistry
1 answer:
blsea [12.9K]3 years ago
4 0

Answer:

In H2CO3(aq) + H2O(l) + CO2(g) there are:

4 hydrogen atoms

2 carbon atoms

6 oxygen atoms

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storchak [24]

Answer:

Choice 4 easy

Explanation:

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How many atoms are in c8h8o3?<br> Show your work
Gelneren [198K]

Answer:

im positive that it's 19

8 0
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If hydrochloric acid has a [H+] of 1.2 x 10-2 M, what is the pH?<br> 1.9<br> 0 1<br> 1.2<br> O<br> 8
Vanyuwa [196]

Answer:

<h2>1.9</h2>

Explanation:

The pH of a solution can be found by using the formula

pH = - log [ {H}^{+} ]

From the question we have

pH =  -  log(1.2 \times  {10}^{ - 2} )  \\  = 1.920818...

We have the final answer as

<h3>1.9</h3>

Hope this helps you

7 0
3 years ago
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A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of
vlada-n [284]

Answer: The molecular of the compound is, C_2H_3O

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=3.095g

Mass of H_2O=1.902g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.095g of carbon dioxide, \frac{12}{44}\times 3.095=0.844g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 1.902g of water, \frac{2}{18}\times 1.092=0.121g of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = (1.621)-[(0.844)+(0.121)]=0.656g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.041 moles.

For Carbon = \frac{0.0703}{0.041}=1.71\approx 2

For Hydrogen  = \frac{0.121}{0.041}=2.95\approx 3

For Oxygen  = \frac{0.041}{0.041}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is C_2H_3O_1=C_2H_3O

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{46.06}{43}=1

Molecular formula = (C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O

Therefore, the molecular of the compound is, C_2H_3O

6 0
3 years ago
A scientist triples the temperature of a gas at constant pressure. By what factor does the gas’s volume change?
laila [671]
Hey there,

A scientist triples the temperature of a gas at constant pressure. By what factor does the gas’s volume change?

I believe that  when the scientist triples the temperature of the gas at a certain and constant pressure, the factor of the gas volume would change most likely with the double amount of pressure that the scientist put in the first place.

~Jurgen
4 0
3 years ago
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