All categories

3 years ago

4. Energy can be conserved by -

Chemistry

1 answer:

mart [117]3 years ago

4 0

Energy can be conserved by efficient energy use.

Answer: Option A

<u>Explanation:</u>

Energy can be transferred from one form to another, but it cannot be destroyed or created. So it can be conserved if efficiently used. Thus efficient usage of energy lead to conservation of energy. Due to conservation of energy, the forces can be renewable and non-renewable.

So, we should know how the input energy can be completely converted to another form of energy leading to efficient usage of energy without any loss. As if there is no loss, input energy will be equal to output energy leading to 100% efficiency.

You might be interested in

The taste of drinking water can be improved by removing impurities from our municipal water by adding substances to the water th

tia_tia [17]

Abcdefghigklmnopqrstuvwxyandz

3 0

3 years ago

How many double bonds are in lewis structure of oxygen difluoride, OF2? 1,0, 2, or 3

solmaris [256]

Answer:

Explanation:

An image of the lewis structure of the compound OF2 is shown in the image attached.

A Lewis structure is a structure in which electron pairs on atoms are shown as dots. Sometimes shared electron pairs are shown by a horizontal straight line connecting the two atoms involved.

OF2 has no double bonds as shown in its structure. It is a compound containing only two O-F sigma bonds and no pi-bonds.

5 0

3 years ago

Which mass of magnesium sulphate will be formed if 12 g of magnesium are reacted with sulphuric acid?

Alinara [238K]

Answer:

magnesium hydrosulphate

4 0

3 years ago

The terms Q and K refer to reaction components at non-equilibrium and equilibrium conditions, respectively. For a forward reacti

daser333 [38]

Answer:

The value of Q must be less than that of K.

Explanation:

The difference of K and Q can be understood with the help of an example as follows

A ⇄ B

In this reaction A is converted into B but after some A is converted , forward reaction stops At this point , let equilibrium concentration of B be [B] and let equilibrium concentration of A be [A]

In this case ratio of [B] and [A] that is

K = [B] / [A] which is called equilibrium constant.

But if we measure the concentration of A and B ,before equilibrium is reached , then the ratio of the concentration of A and B will be called Q. As reaction continues concentration of A increases and concentration of B decreases. Hence Q tends to be equal to K.

Q = [B] / [A] . It is clear that Q < K before equilibrium.

If Q < K , reaction will proceed towards equilibrium or forward reaction will

proceed .

8 0

3 years ago

7. The equilibrium constant Kc for the reaction H2(g) + I2(g) ⇌ 2 HI(g) is 54.3 at 430°C. At the start of the reaction there are

Juli2301 [7.4K]

Answer:

[H2] = 0.0692 M

[I2] = 0.182 M

[HI] = 0.826 M

Explanation:

Step 1: Data given

Kc = 54.3 at 430 °C

Number of moles hydrogen = 0.714 moles

Number of moles iodine = 0.984 moles

Number of moles HI = 0.886 moles

Volume = 2.40 L

Step 2: The balanced equation

H2 + I2 → 2HI

Step 3: Calculate Q

If we know Q, we know in what direction the reaction will go

Q = [HI]² / [I2][H2]

Q= [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Q =(n(HI)²) /(nH2 *nI2)

Q = 0.886²/(0.714*0.984)

Q =1.117

Q<Kc This means the reaction goes to the right (side of products)

Step 2: Calculate moles at equilibrium

For 1 mol H2 we need 1 mol I2 to produce 2 moles of HI

Moles H2 = 0.714 - X

Moles I2 = 0.984 -X

Moles HI = 0.886 + 2X

Step 3: Define Kc

Kc = [HI]² / [I2][H2]

Kc = [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Kc =(n(HI)²) /(nH2 *nI2)

KC = 54.3 = (0.886+2X)² /((0.714 - X)*(0.984 -X))

X = 0.548

Step 4: Calculate concentrations at the equilibrium

[H2] = (0.714-0.548) / 2.40 = 0.0692 M

[I2] = (0.984 - 0.548) / 2.40 = 0.182 M

[HI] = (0.886+2*0.548) /2.40 = 0.826 M

6 0

3 years ago

Read 2 more answers