Someone people help me with this chemistry question

Chemistry

1 answer:

o-na [289]2 years ago

6 0

Answer:

6 moles

Explanation:

From the equation you can see that twice as many HCL moles are used as H2 produced

2 x 3 = 6 moles of HCL required

You might be interested in

In the diagram, which letter represents the transition from liquid to gas?<br> B<br> A<br> D<br> C

lana66690 [7]

I got on here because I don't understand the question but I did my best to answer because I noticed you asked 3 days ago. IF I'm right the answer is D. My diagram shows
A at -50 °C
B at 0 °C
C at 50 °C
D at 100 °C (gas to liquid or liquid to gas)
And E at 150 °C
So I hope I'm right because I'm answering the same question.

5 0

3 years ago

In which molecule the attraction between H and the atom to which it is bonded be the greatest

jolli1 [7]

I got the one you got to pick up my car stereo truck in my truck so I’m coming

5 0

3 years ago

Calculate the molar solubility of PbBr2 (Ksp = 4.67x10-6) in 0.10M NaBr solution.

Softa [21]

Explanation:

$PbBr_{2}$ will dissociate into ions as follows.

$PbBr_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^{-}(aq)$

Hence, $K_{sp}$ for this reaction will be as follows.

$K_{sp} = [Pb^{2+}][Br^{-}]^{2}$

We take x as the molar solubility of $PbBr_{2}$ when we dissolve x moles of solution per liter.

Hence, ionic molarities in the saturated solution will be as follows.

$[Pb^{2+}]$ = $[Pb^{2+}]_{o}$ + x

$[Br^{-}]^{2}$ = $[Br^{-}]_{o}$ + 2x

So, equilibrium solubility expression will be as follows.

$K_{sp}$ = $([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}$

Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains $[Br^{-}]_{o}$ = 0.10 and there will be no lead ions. So, $[Pb^{2+}]_{o}$ = 0

So, $[Br^{-}]_{o} + 2x$ will approximately equals to $[Br^{-}]_{o}$ .