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Sati [7]
3 years ago
6

Using the combined gas law above, identify the variables that would be in the numerator (A) and denominator (B) if you were to r

earrange the gas law to solve for final temperature.
T subscript 2 equals StartFraction A over B EndFraction.

Denominator (B):

P1V1T2

P2V2T1

T1V2

P1V1
Chemistry
1 answer:
ad-work [718]3 years ago
4 0

Answer:

The denominator B = P1V1

Explanation:

We'll begin by writing the combine gas equation. This is shown below:

P1V1/T1 = P2V2/T2

Where:

P1 is the initial pressure.

V1 is the initial volume.

T1 is the initial temperature.

P2 is the final pressure.

V2 is the final volume.

T2 is the final temperature.

Now let us make T2 the subject of the formula to obtain our desired result. This is illustrated below:

P1V1/T1 = P2V2/T2

Cross multiply to express in linear form

P1V1T2 = P2V2T1

Divide both side by P1V1

T2 = P2V2T1/P1V1

From the above illustration,

The numerator A = P2V2T1

The denominator B = P1V1

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saw5 [17]

Yo sup??

we can solve this problem by applying Newton's 2nd law

F*t=Δp

p=momentum

pi=mu=1500*30

pf=mv=m*0=0

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5 0
3 years ago
Which radioactive emission has the smallest (least) mass? Question options:
Tems11 [23]

Answer:

The answer is D. gamma rays

Explanation:

A radioactive atom can have three different types of emission:

alpha particles (α) = they have a mass of 4 amu and they have a very low penetrating power.

Beta particles (β) = they have 5x10^{-4} amu and they have an intermediate penetrating power

Gamma rays (γ) = they are not particles basically just energy so its mass is ≈ 0 and its penetrating power is higher

For this reason Gamma emissions (γ) has the smallest mass value.

8 0
3 years ago
What mass of chromium would be produced from the reaction of 57.0 g of potassium with 199 g of chromium(II) bromide according to
fredd [130]

Answer:

Mass of Chromium produced = 37.91 grams

Explanation:

2K + CrBr₂  →  2KBr + Cr

2mole     1 mole                1 mole

mass of Potassium = 57.0 grams

molar mass of Potassium = 39.1 g/mol

no of moles of Potassium = 57.0 / 39.1 = 1.458 moles

mass of CrBr₂= 199 grams

molar mass of CrBr₂ = 211.8 gram/mole

no of moles of CrBr₂ = 199 / 211.8 = 0.939 mole

From chemical equation

1 mole of CrBr₂ = 2 moles of K

∴ 0.939 moles of CrBr₂ = ?

   ⇒ 0.939 x 2/1 = 1.878 moles of K

1.878 moles of K is needed, but there is 1.458 moles of K. So, Potassium is completed first during the reaction . Hence, Potassium is limiting reagent. and CrBr₂ is excess reagent .

From chemical equation

2 moles of K = 1 mole of Cr

∴ 1.458 moles of K = ?

   ⇒ 1.458 x 1/ 2 = 0.729 moles of Cr

no of moles of Cr formed = 0.729 moles

molar mass of Cr = 52.0 g/mol

mass of one mole of Cr = 52.0 grams

mass of 0.729 moles of Cr = 52.0 x 0.729 = 37.908 grams

mass of Chromium produced = 37.91 grams

6 0
3 years ago
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Allushta [10]

Answer:

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hope this help plz give me a thxs and brainlyest and a five star thx and you welcome let me know if this helped

Explanation:

4 0
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Daniel [21]

Answer:

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Explanation:

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