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3 years ago

Which statement best describes the law of conservation of energy?

1 answer:

8 0

Correct answer - B (The total amount of energy does not change during an energy conversion.)

Why? - The total amount of energy and matter in the Universe remains constant, merely changing from one form to another. The First Law of Thermodynamics (Conservation) states that energy is always conserved, it cannot be created or destroyed. In essence, energy can be converted from one form into another.

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A student makes two mixtures, both made of the same substances. What must be true of the two mixtures? A Both mixtures have the

Goshia [24]

Answer:

im not completely sure but the mst likely answer would be the would be they have the same density

Explanation:

if two mixture have the same ingriedients their most likely going to have the same density depending of the measurements like when baking a cake when you add your wet ingriedients to your dry one it makes a batter and if you were to make another mixture with the same ingriedients and somewhat similar measurements your going to get a similar density

6 0

3 years ago

The following information is given for benzene, C6H6, at 1atm: boiling point = 80.1 °C Hvap(80.1 °C) = 30.7 kJ/mol specific heat

user100 [1]

<u>Answer:</u> The heat required for the process is 4.24 kJ

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of benzene = 24.8 g

Molar mass of benzene = 78.11 g/mol

Putting values in above equation, we get:

$\text{Moles of benzene}=\frac{24.8g}{78.11g/mol}=0.318mol$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

$q$ = amount of heat absorbed = ?

n = number of moles = 0.318 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = 30.7 kJ/mol

Putting values in above equation, we get:

$30.7kJ/mol=\frac{q}{0.318mol}\\\\q=(30.7kJ/mol\times 0.318mol)=4.24kJ$

Hence, the heat required for the process is 4.24 kJ

6 0

3 years ago

Molecule contains carbon, hydrogen and sulfur atoms. When a sample of 0.535g of this compound is burnt in oxygen, 1.119 g of CO2

OLga [1]

Answer:

The empirical formula is, C4H4S

Explanation:

Number of moles of carbon = 1.119 g/ 44g/mol = 0.025 moles

Mass of Carbon= 0.025 moles × 12 g/ mole = 0.3 g

Number of moles of hydrogen = 0.229/18g/mol × 2 = 0.025 moles

Mass of hydrogen = 0.025 moles × 1 = 0.025 g

Number of moles of sulphur = 0.407g/ 64 g/mol = 0.0064 moles

Mass of sulphur= 0.0064 moles ×32 = 0.2 g

Now we obtain the mole ratios by dividing through by the lowest ratio.

C- 0.025 moles/ 0.0064 moles, H- 0.025 moles/ 0.0064 moles, S- 0.0064 moles/0.0064 moles

C4H4S

4 0

2 years ago

Which of the following is not an example of a chemical change?

elena55 [62]

This would be C. A car rusting

5 0

3 years ago

Read 2 more answers

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