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jekas [21]
3 years ago
5

The atomic number of carbon is 6. what is the mass number of a carbon atom with 5 neutrons?

Chemistry
1 answer:
zhannawk [14.2K]3 years ago
4 0

Atomic number of carbon = Number of protons = 6

Mass number of carbon = Number of protons + neutrons

= 6 + 5 = 11

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Consider the following reaction where K. = 9.52 10 2 at 350 K.
Harman [31]

Answer:

The correct answer is A :))

3 0
2 years ago
Identify which two compounds below are constitutional isomers
Romashka [77]

Answer:

(CH₃)₃COCH3₃ and (CH₃)₂CHOCH₂CH₃

Explanation:

Isomers are compounds which have the same molecular formula. Constitutional isomers have different connectivity; the atoms are connected in different ways.

1. (CH₃)₃COCH₃

2. (CH₃)₂CHOCH3₃

3. (CH₃)₂CHOCH₂CH₃

Molecules 1 and 3 have the same formula (C₅H₁₂O) and are isomers. Molecule 2 is not an isomer. From the structural formula, it is clear that Molecules 1 and 3 have different connectivity.

4 0
3 years ago
B) How many kilograms of carbon dioxide are formed when 24.42 g of iron is<br> produced?
charle [14.2K]

Answer:

0.0289 kg of CO2 will be formed

Explanation:

Step 1: Data given

Mass of iron produced = 24.42 grams

Atomic mass iron = 55.845 g/mol

Molar mass CO2 = 44.01 g/mol

Step 2: The balanced equation

Fe2O3 + 3CO → 2Fe + 3CO2

Step 3: Calculate moles iron

Moles iron = mass iron / molar mass iron

Moles iron = 24.42 grams / 55.845 g/mol

Moles iron = 0.437 moles

Step 4: Calculate moles CO2

For 1 mol Fe2O3 we need 3 moles CO to produce 2 moles Fe and 3 moles CO2

For 0.437 moles Fe we'll have 3/2 * 0.437 = 0.6555 moles CO2

Step 5: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.6555 moles * 44.01 g/mol

Mass CO2 = 28.85 grams = 0.0289 kg

0.0289 kg of CO2 will be formed

5 0
3 years ago
An alloy is made up of more than one type of what?
pychu [463]
More than one type of metallic elements.
4 0
3 years ago
What factors affect the dynamic state of equilibrium in a chemical reaction and how?
yanalaym [24]

Answer:

Only changes in temperature will influence the equilibrium constant K_c. The system will shift in response to certain external shocks. At the new equilibrium Q will still be equal to K_c, but the final concentrations will be different.

The question is asking for sources of the shocks that will influence the value of Q. For most reversible reactions:

  • External changes in the relative concentration of the products and reactants.

For some reversible reactions that involve gases:

  • Changes in pressure due to volume changes.

Catalysts do not influence the value of Q. See explanation.

Explanation:

\displaystyle K_c = {e}^{\Delta G/(R\cdot T)}.

Similar to the rate constant, the equilibrium constant K_c depends only on:

  • \Delta G the standard Gibbs energy change of the reaction, and
  • T the absolute temperature (in degrees Kelvins.)

The reversible reaction is in a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. Reactants are constantly converted to products; products are constantly converted back to reactants. However, at equilibrium Q = K_c the two processes balance each other. The concentration of each species will stay the same.

Factors that alter the rate of one reaction more than the other will disrupt the equilibrium. These factors shall change the rate of successful collisions and hence the reaction rate.

  • Changes in concentration influence the number of particles per unit space.
  • Changes in temperature influence both the rate of collision and the percentage of particles with sufficient energy of reaction.

For reactions that involve gases,

  • Changing the volume of the container will change the concentration of gases and change the reaction rate.

However, there are cases where the number of gases particles on the reactant side and the product side are equal. Rates of the forward and backward reaction will change by the same extent. In such cases, there will not be a change in the final concentrations. Similarly, catalysts change the two rates by the same extent and will not change the final concentrations. Adding noble gases will also change the pressure. However, concentrations stay the same and the equilibrium position will not change.

8 0
3 years ago
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