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Pani-rosa [81]
3 years ago
7

Describe the temperature changes that occur in ice as energy is added, starting in the frozen state and ending in the vapor stat

e.
Chemistry
1 answer:
Ket [755]3 years ago
8 0

Explanation:

When water is frozen then it is known as ice and its state is solid. So, its molecules will be held closer to each other as they are held by strong intermolecular forces of attraction.

As a result, its temperature will be minimum as its molecules have least kinetic energy.

It is known that kinetic energy of a substance is directly proportional to temperature.

As,            K.E = \frac{3}{2}kT

where            K.E = kinetic energy

                       T = temperature

                       k = boltzmann constant

When solid changes into liquid state then it means molecules of a substance has gained kinetic energy due to which there occurs more collisions between the molecules.

Hence, temperature of substance also increases.

Whereas when liquid state of a substance changes intro vapor state then it means that more kinetic energy has gained by the molecules due to which there will be much more collisions between the molecules.

Hence, temperature will be maximum in vapor state.

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Name the following alkyne: CH3CH₂C = CCH₂CH₂CH3
liubo4ka [24]

The given alkyne is Option A 3-heptyne

<h3>What is an Alkyne ?</h3>

The hydrocarbon having at least one C-C triple bond is called an Alkyne.

It has the general formula of   \rm C_n H_{2n+2} .

In the question it is being mentioned that it is an alkyne so there will be a triple bond and not a double bond.

It has been asked in the question that

CH3CH₂C ≡  CCH₂CH₂CH3 is which alkyne from the given option.

The counting of the Carbon chain is done from the left side and the Triple bond is at the 3rd Carbon , so 3-heptyne .

To know more about Alkyne

brainly.com/question/23508203

#SPJ1

<h3 />

6 0
1 year ago
1Which is a dopant for a p-type semiconductor?
aalyn [17]

For a p type of semiconductor we need a dopant which is from 13th group in periodic table

Al , B, Ga, In Tl

So the correct element will be In : Indium

The other elements belongs to 15th group and hence will give n type semiconductor


5 0
3 years ago
Read 2 more answers
Select the correct answer.
Svetlanka [38]

Answer:

alcohol

I hope this helps you

7 0
3 years ago
Read the following chemical equation
Naddika [18.5K]

Answer:

For eacht 4 moles Fe consumed, we will produce 2 moles Fe2O3

The mole ration is 4:2 (option 1)

Explanation:

Step 1: The unbalanced equation

Fe + O2 → Fe2O3

Step 2: Balancing the equation

Fe + O2 → Fe2O3

On the left side we have 2x O (in O2) and on the right side we have 3x O (in Fe2O3) . To balance the amount of O on both sides, we have to multiply O2 by 3 and Fe2O3 by 2.

Fe + 3O2 → 2Fe2O3

On the left side we have 1x Fe, On the right side we have 4x (in 2Fe2O3). To balance the amount of Fe we have to multiply Fe (on the left side) by 4.

Now the equation is balanced.

4Fe + 3O2 → 2Fe2O3

For eacht 4 moles Fe consumed, we will produce 2 moles Fe2O3

The mole ration is 4:2 (option 1)

8 0
3 years ago
Calculate the pH of 1.00 L of a buffer that contains 0.105 M HNO2 and 0.170 M NaNO2. What is the pH of the same buffer after the
kari74 [83]

Answer:

1.- pH =3.61

2.-pH =3.53

Explanation:

In the first part of this problem we can compute the pH of the buffer by making use of the Henderson-Hasselbach equation,

pH = pKa + log [A⁻]/[HA]

where [A⁻] is the conjugate base anion concentration ( [NO₂⁻]), [HA] is the weak acid concentration,[HNO₂].

In the second part, our strategy has to take into account that some of the weak base NO₂⁻ will be consumed by reaction with the very strong acid HCl. Thus, first we will calculate the new concentrations, and then find the new pH similar to the first part.

First Part

pH  = 3.40+ log {0.170 /0.105}

pH =  3.61

Second Part

# mol HCl = ( 0.001 L ) x 12.0 mol / L = 0.012

# mol NaNO₂ reacted = 0.012 mol ( 1: 1 reaction)

# mol NaNO₂ initial = 0.170 mol/L x 1 L = 0.170 mol

# mol NaNO₂ remaining = (0.170 - 0.012) mol  = 0.158

# mol HNO₂ produced = 0.012 mol

# mol HNO₂ initial = 0.105

# new mol HNO₂ = (0.105 + 0.012) mol = 0.117 mol

Now we are ready to use the Henderson-Hasselbach with the new ration. Notice that we dont have to calculate the concentration (M) since we are using a ratio.

pH = 3.40 + log {0.158/.0117}

pH = 3.53

Notice there is little variation in the pH of the buffer. That is the usefulness of buffers.

4 0
2 years ago
Read 2 more answers
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