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LenaWriter [7]
3 years ago
14

Study the image. Warm and cold air fronts with points labeled 1 through 4. 1: cold air entering from the left and moving down. 2

: clouds with rain. 3: warm air entering from the right and moving up. 4: warm air entering from the top left and moving down toward the clouds. Warm and cold air fronts with points labeled 1 through 4. 1: cold air entering from the left and moving down. 2: clouds with rain. 3: warm air entering from the right and moving up. 4: warm air entering from the top left and moving down toward the clouds. At which point is cool air circulating beneath warm air? 1 2 3 4
Chemistry
2 answers:
nata0808 [166]3 years ago
7 0

Answer:

1

Explanation:

NISA [10]3 years ago
4 0

Answer:

1

Explanation:

Edge 2021

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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
Hydrogen gas H2 is the least dense of all gases. A sample of H2 gas is found to occupy a volume of 1.23 L at 755 mmHg and 0 ◦C.
lana66690 [7]

Answer:

V₂ = 1.48 L

Explanation:

Given data:

Initial volume = 1.23 L

Initial pressure = 755 mmHg (755/760 =0.99 atm)

Initial temperature = 0 °C (0 +273 = 273 K)

Final temperature = 50°C (50+273 = 323 K)

Final volume =  ?

Final pressure = 0.97 atm

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 0.99 atm × 1.23 L × 323 K / 273 K × 0.97 atm

V₂ = 393.32 atm .L. K / 264.81 atm.K

V₂ = 1.48 L

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3 years ago
How many valence electrons are in the electron-dot structures for the elements in group 3a(13)?
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Three valence electrons
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