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Art [367]
2 years ago
8

Which of the following is true for a chemical reaction at equilibrium?

Chemistry
1 answer:
victus00 [196]2 years ago
7 0

Answer:

youre gonna have to include the answers for me to help

Explanation:

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Plz help me guys<br><br> A<br><br> B<br><br> C<br><br> D
Taya2010 [7]

Answer:

C

Explanation:

I think

3 0
3 years ago
Describe how nucleic acid basis pair up
castortr0y [4]

The rules of base pairing (or nucleotide pairing) are: A with T: the purine adenine (A) always pairs with the pyrimidine thymine (T) C with G: the pyrimidine cytosine (C) always pairs with the purine guanine (G)

The nucleotides in a base pair are complementary which means their shape allows them to bond together with hydrogen bonds. The A-T pair forms two hydrogen bonds. The C-G pair forms three. The hydrogen bonding between complementary bases holds the two strands of DNA together.

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
What do you understand by shared pair of electron ​
PilotLPTM [1.2K]

Answer:Lone pairs are the valence electron pair of any element which do not take part in bonding but affect the shape of molecules. Bond pairs or shared pairs are the electron pair which does both affect the geometry of molecules and take part in chemical bonding. These are form due to sharing of electrons.

Explanation:

5 0
3 years ago
If 21.3 grams of lithium react with excess water, how many liters of hydrogen gas can be produced at 297 Kelvin and 1.40 atmosph
Gnoma [55]
Hope it helped you.

4 0
3 years ago
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