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Verizon [17]
2 years ago
15

What is the name of the functional group that is attached to this hydrocarbon?

Chemistry
1 answer:
Ksenya-84 [330]2 years ago
5 0

Answer: alkyl halide

Explanation:

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If iron metal reacts with an aqueous solution of silver nitrate and zinc reacts with an aqueous solution of iron sulfate, rank t
Lelechka [254]

Answer:

yes!you are right a cloudy formation will be formed when they will react.its because if nitrogen.

5 0
3 years ago
If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00 atm, what will be its
yKpoI14uk [10]

Answer:

1.44 L

Explanation:

If the temp does not change

P1V1 = P2V2

1 * 3.6   = 2.5 * V2

V2 = 1.44 L

5 0
2 years ago
Why do the elements in a group all behave similarly
Svetradugi [14.3K]

Answer:

See the explanation below, please.

Explanation:

The elements of the periodic table that belong to the same group (each column) have similar physical and chemical properties. This is because they have the same number of electrons in their last electronic layer.

Example of electronic configuration of elements of GROUP IA:

Hydrogen: 1s ^ 1

Lithium: 1s ^ 2 2s ^ 1

6 0
3 years ago
Balance the following equation. Then, given the moles of reactant or product below, determine the corresponding amount in moles
Mrrafil [7]

Answer:

Option a → 4 mol NH₃

Explanation:

This the unbalanced reaction

NH₃  +  O₂  ⟶  N₂  +  H₂O

The balanced reaction:

4NH₃  +  3O₂  →  2N₂  +  6H₂O

4 mol of ammonia

3 mol of oxygen

2 mol of nitrogen

6 mol of water

6 0
3 years ago
What is the entropy change in the environment when 5.0 MJ of energy is transferred thermally from a reservoir at 1000 K to one a
Leni [432]

Answer:

The entropy change in the environment is 3.62x10²⁶.

Explanation:

The entropy change can be calculated using the following equation:

\Delta S = \frac{Q}{k_{B}}(\frac{1}{T_{f}} - \frac{1}{T_{i}})

Where:

Q: is the energy transferred = 5.0 MJ

k_{B}: is the Boltzmann constant = 1.38x10⁻²³ J/K  

T_{i}: is the initial temperature = 1000 K

T_{f}: is the final temperature = 500 K

Hence, the entropy change is:

\Delta S = \frac{5.0 \cdot 10^{6} J}{1.38 \cdot 10^{-23} J/K}(\frac{1}{500 K} - \frac{1}{1000 K}) = 3.62 \cdot 10^{26}                                    

Therefore, the entropy change in the environment is 3.62x10²⁶.

I hope it helps you!          

7 0
3 years ago
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