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3 years ago

13

the volume of a gas with a pressure of 1.2 atmospheres increases from 1 to 4 what is the final pressure of the gas assuming cons

tant temperature

1 answer:

Oksana_A [137]3 years ago

7 0

Answer:

the answer is is putting ptessure on the gas

Explanation:

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In the following pair, determine whether the two represent resonance contributors of a single species or depict different substa

Aleksandr-060686 [28]

Answer:

They are resonance contributors

Explanation:

Resonance structures are structures that differ only in the distribution or placement of electrons.

Considering the two structures, we can easily see that the two species have the same total number of bonds and electrons differing only in the distribution of these electrons.

Hence, they are resonance contributors.

5 0

3 years ago

Please help me out...

liq [111]

Answer:

Explanation:

To convert from grams to atoms, first divide by the molar mass, the multiply by 6.022*10^23.

To convert from moles to mass, multiply by the molar mass of the element.

Hope this helps!

-Emma Victoria

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2 years ago

The specific heat of aluminum is 0.897 J/g•°C. Which equation would you use to calculate the amount of heat needed to raise the

docker41 [41]

Q = 0.75 g x 0.897 J/g•°C x 22°C

6 0

3 years ago

I am a little confused

Sonbull [250]

The answer is D. I did that and i got it right.

7 0

3 years ago

Read 2 more answers

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2

Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

$Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}$

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

$volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL$

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

7 0

3 years ago