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aivan3 [116]
3 years ago
11

Please help! (The question is in the picture above)

Chemistry
1 answer:
goldfiish [28.3K]3 years ago
7 0
(i) The pH of HCl is between 0 and 2.
(ii) the presence of Carbon Dioxide turns lime water milky or sees a white precipitate form on the surface of the lime water.  Lime water is CaOH and when it reacts with Carbon Dioxide it forms a white solid (precipitate) Calcium Carbonate.
Ca(OH)_{2}  _{(l)}   +   CO_{2} _{(g)}  -----\ \textgreater \   CaCO_{3} _{(s)}  +  H_{2}O_{(l)} 
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7 0
3 years ago
PLEASE HELP ME I NEED HELP. DO NOT GIVE ANY WRONG ANSWERS.
Lemur [1.5K]
I believe the answer is C
3 0
3 years ago
A tank of oxygen with a pressure of 23 atm is moved from room temperature of 293K to a storage freezer at 239K. What is the fina
Gelneren [198K]

Answer:

18.76atm

Explanation:

Using the formula V1P1/T1 = V2P2/T2, from combined gas law. Volume is constant since we have not been given. Therefore the formula comes to be; P1/T1 = P2/T1

To get P2 = T2(P1/T1)

Where P2 is final pressure

P2 = 239K ( 23atm/293K)

=18.76atm

7 0
3 years ago
When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular
Luda [366]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is C_2H_2O_4 and C_6H_4O_2

<u>Explanation:</u>

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=1.03g

Mass of H_2O=0.14g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

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

So, in 1.03 g of carbon dioxide, \frac{12}{44}\times 1.03=0.28g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

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

So, in 0.14 g of water, \frac{2}{18}\times 0.14=0.016g of hydrogen will be contained.

  • Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles

  • <u>Step 2:</u> 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 0.00775 moles.

For Carbon = \frac{0.023}{0.00775}=2.96\approx 3

For Hydrogen  = \frac{0.016}{0.00775}=2.06\approx 2

For Oxygen  = \frac{0.00775}{0.00775}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is C_3H_{2}O_1=C_3H_2O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

n=\frac{108.10g/mol}{54g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2

Thus, the empirical and molecular formula for the given organic compound is C_3H_2O and C_6H_4O_2

6 0
3 years ago
Please help noowwwww... if you don’t know please don’t answer
pshichka [43]

Answer:

i honetly dont thing anyone knows that so look it up

Explanation:

7 0
3 years ago
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