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3 years ago

Can renewable energy be used again and again?

Chemistry

1 answer:

Feliz [49]3 years ago

3 0

Answer:

Even renewable resources can be used unsustainably. We can cut down too many trees without replanting. We might need grains for food rather than biofuels. Some renewable resources are too expensive to be widely used

Explanation:

hope this helps

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The following reaction is at equilibrium in a sealed container.

ale4655 [162]

Answer:

D.Lowering the temperature is the best option.

Explanation:

The value of equilibrium constants aren't changed with change in the pressure or concentrations of reactants and products in equilibrium. The only thing that changes the value of equilibrium constant is a change of temperature.

In the reaction below for example;

A + B <==>C+D

If you have moved the position of the equilibrium to the right (and so increased the amount of C and D), why hasn't the equilibrium constant increased?

Let's assume that the equilibrium constant mustn't change if you decrease the concentration of C - because equilibrium constants are constant at constant temperature. Why does the position of equilibrium move as it does?

If you decrease the concentration or pressure of C, the top of the Kc expression gets smaller. That would change the value of Kc. In order for that not to happen, the concentrations of C and D will have to increase again, and those of A and B must decrease. That happens until a new balance is reached when the value of the equilibrium constant expression reverts to what it was before.

3 0

3 years ago

Characteristics used to describe matter are called

Shtirlitz [24]

Answer:

A.physical properties

5 0

2 years ago

When calcium carbonate is heated, it decomposes to produce calcium oxide and carbon dioxide, as shown in the diagram below. How

Vikki [24]

Explanation:

Before we start with any calculation, we need to determine the stoichiometric relationship between the reactants consumed and the products formed. In other words, we need to establish a balanced chemical equation that describes the decomposition of calcium carbonate as shown below.

$\text{CaCO}_{3} \ \ \text{(s)} \ \overset{\Delta}{\longrightarrow} \ \text{CaO} \ \text{(s)} \ + \ \text{CO}_{2} \ \text{(g)}$ ,

where the uppercase delta symbol, $\Delta$ , above the chemical reaction arrow denotes that heat is applied to make the reaction proceed in the direction of the arrow.

To calculate how many moles are there in 4 grams of calcium carbonate, we use the relation

$n \ = \ \displaystyle\frac{m}{M_{r}}$ ,

where $n$ is the number of moles, $m$ is the mass of the chemical substance and $M_{r}$ is the molar mass of the chemical substance. We first need to identify the molar mass of calcium carbonate,

$M_{r} \, [\text{CaCO}_{3}] \ = \ A_{r} \, [\text{Ca}] \ + \ A_{r} \, [\text{C}] \ + \ 3 \, \times \, A_{r} \, [\text{O}] \\ \\ M_{r} \, [\text{CaCO}_{3}] \ = \ 40.078 \ + \ 12.011 \ + \ 3 \, \times \, 15.999 \\ \\ M_{r} \, [\text{CaCO}_{3}] \ = \ 100.086 \ \text{g mol}^{-1}$

Hence,

$n \ = \ \displaystyle\frac{4 \ \text{g}}{100.086 \ \text{g mol}^{-1}} \\ \\ \\ n \ = \ 0.04 \ \text{mol}$ .

We know that, from the balanced chemical equation above, 1 mole of calcium carbonate following decomposition, will theoretically lead to the production of 1 mole of carbon dioxide. Then, if 0.04 moles of calcium carbonate is decomposed, then, 0.04 moles of carbon dioxide is produced.

To convert the number of moles into the mass of carbon dioxide produced, we need to rearrange the formula as follows.

$n \ = \ \displaystyle\frac{m}{M_{r}} \\ \\ \\m \ = \ n \ \times \ {M_{r}}$ .

Thus,

$M_{r} \, [\text{CO}_{2}] \ = \ A_{r} \, [\text{C}] \ + \ 2 \, \times \, A_{r} \, [\text{O}] \\ \\ M_{r} \, [\text{CO}_{2}] \ = \ (12.011 \ + \ 2 \, \times \, 15.999) \ \text{g mol}^{-1} \\ \\ M_{r} \, [\text{CO}_{2}] \ = \ 44.009 \ \text{g mol}^{-1}$