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snow_lady [41]
3 years ago
14

Which of the following is a measurement conversion that you may need to make when comparing energy and mass in nuclear reactions

?
Chemistry
2 answers:
sergey [27]3 years ago
5 0

Answer:be

Explanation:because the comparing energy have more than mass in nuclear

Sloan [31]3 years ago
5 0

Answer:

Kilojoules to joules

Explanation:

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How could you use these solutions to determine the identities of each metal powder?
Anastaziya [24]

Answer:

1.) The nitric acid solution will oxidize and thus dissolve _*(Zn and Pb)*_. This will allow to identify _**Pt**_.

2) To distinguish between _*(Zn and Pb)*_, we can use the nickel nitrate.

3) The nickel nitrate solution will oxidize and thus dissolve _**Zn**_ and will not oxidize or dissolve _**Pb**_.

Explanation:

1) Unlike Zinc and Lead, Platinum does not react with Nitric acid. So, it will be the only metal from step 1 that doesn't react. Pt is identified in this manner.

2) Nickel is higher than Lead in the activity series, but Zinc is higher than both of them in the activity series. This selectivity can be used to distinguish between Zinc and Lead metal powders.

3) Because Zinc is higher than Nickel in the activity series, it means that Zinc metal can and will displace Nickel from Nickel Nitrate solution. Therefore the Nickel Nitrate solution will oxidize and dissolve the Zinc metal.

But, there will be no reaction with the Lead metal powders sample as Pb is lower than Ni in the activity series, so, Nickel Nitrate solution will not oxidize or dissolve the Lead metal powders.

5 0
3 years ago
An isotope contains 26 protons, 24 electrons, and 32 neutrons.
lisov135 [29]
26
protons equal identity
6 0
3 years ago
Read 2 more answers
The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air
vagabundo [1.1K]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

(c) Determine the final temperature of the air in the room after the combustion.

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 =  5.56 ×10⁴  × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴  × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ =  ΔT +  T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

7 0
4 years ago
Choose two different particles found in the air. Explain how they are different and how they are the same
BARSIC [14]

Answer:

Dust and smoke.

Explanation:

Dust and smoke are two different particles present in the air. Dust and smoke are different from one another due to their origin. Smoke formed from burning of materials while dust refers to the soil particles lifted by the wind due to their light weight. Dust and smoke are similar to each other due to their small in size, infinite number means uncountable and light weight.

6 0
3 years ago
Describe the green house effect
eimsori [14]

Answer:

The greenhouse effect is the process by which radiation from a planet's atmosphere warms the planet's surface to a temperature above what it would be without this atmosphere. Radiatively active gases in a planet's atmosphere radiate energy in all directions.

7 0
3 years ago
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