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.
Answer: S,Fe,O2,Sr, and Al
Explanation: I saw it on quizlet
This dilution problem uses the equation
M
a
V
a
=
M
b
V
b
M
a
= 6.77M - the initial molarity (concentration)
V
a
= 15.00 mL - the initial volume
M
b
= 1.50 M - the desired molarity (concentration)
V
b
= (15.00 + x mL) - the volume of the desired solution
(6.77 M) (15.00 mL) = (1.50 M)(15.00 mL + x )
101.55 M mL= 22.5 M mL + 1.50x M
101.55 M mL - 22.5 M mL = 1.50x M
79.05 M mL = 1.50 M
79.05 M mL / 1.50 M = x
52.7 mL = x
59.7 mL needs to be added to the original 15.00 mL solution in order to dilute it from 6.77 M to 1.50 M.
I hope this was helpful.
Answer:
Ok, so the process here is to convert the mass of H2 (hydrogen gas) to moles by dividing the mass by the molar mass of H2. Once you have the moles then you have to multiply by the STP (standard temperature and pressure) molar volume which should be 22.4.
Molar mass of H2 = (1.01)x2 = 2.02g/mol
19.3/2.02 = 9.55 moles
Now just multiply the moles by the molar volume
9.55 moles x 22.4 = 213.92 Litres of H2 are in 19.3g of H2
Thank you for posting your question here. Below is the solution:
HNO3 --> H+ + NO3-
<span>HNO3 = strong acid so 100% dissociation </span>
<span>** one doesn't need to find the molarity of water since it is the solvent </span>
<span>0M HNO3 </span>
<span>1x10^-6M H3O+ </span>
<span>1x10^-6M NO3- </span>
<span>1x10^-8M OH-.....the Kw = 1x10^-14 = [H+][OH-] </span>
<span>you have 1x10^-6M H+ so, 1x10^-14 / 1x10^-6 = 1x10^-8M OH- </span>
<span>1x10^-6 Ba(OH)2 = strong base, 100% dissociation </span>
<span>1x10^-6M Ba2+ </span>
<span>2x10^-6M OH- since there are 2 OH- / 1 Ba2+ </span>
<span>0M Ba(OH)2 </span>
<span>5x10^-9M H3O+</span>
