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omeli [17]
3 years ago
7

4.0 L sample of methane gas is collected at 30.0c, what is the volume of the sample at -8c?

Chemistry
1 answer:
Vedmedyk [2.9K]3 years ago
8 0
This particular law is a gas law, called Charle's Law. The formula is:

V1    V2
---- = ----
T1     T2

So we know our original volume is 4.0L, so we would plug that into our V1. We know T1 is the 30 degrees, since it relates to our original volume. However, we need to convert it to kelvin. We do this simply by adding 273 degrees to the 30 degrees, since 273 is the constant for kelvin.

We do not know our second volume, however we know out T2. It is -8 degrees, and don't forget to convert it to Kelvin. 

So, when we plug all of these numbers into the equation, we are left with V2 to find. To do this we cross multiply (V1 x T2) and then divide by T1. That leaves us with the number for V2. Don't forget to round to the least # of sig figs! And you can divide V1 by T1, and then divide V2 by T2, to ensure your answers are the same, since they are directly porportional and need to be equal to each other. 

Hope I could help!
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In an exothermic reaction, the bonding energy of the product is:
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How to prepared sodium chloride solution in the laboratory.​
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2 years ago
If 15.0 mL of phosphoric acid completely neutralizes 38.5 mL of 0.150 mol/L calcium hydroxide, what is the concentration of the
Sedbober [7]

Answer:

Let me give it a try.

H3PO4 + Ca(OH)2 = Ca3(PO4)2 + H2O

Balancing this reaction

2H3PO4 + 3Ca(OH)2 == Ca3(PO4)2 + 6H2O.

Moles= Molarity x Volume

Volume = 38.5ml = 0.0385L

Moles of Ca hydroxide = 0.150m/L x 0.0385L

(Notice the units canceling out...leaving moles).

=0.005775moles of Ca(OH)2.

From balanced reaction...

3moles of Ca(OH)2 completely reacts with 2moles of H3PO4

0.005775moles of Ca(OH)2 would completely react with....

= 0.005775 x 2/(3)

=0.00385moles of H3PO4.

Now we're looking for its Concentration in Mol/L

Molarity=Moles of solute/Volume of solution(in L)

Volume of solution assuming no other additions to the reaction = 15ml + 38.5ml =53.5ml =0.0535L

Molarity = 0.00385/0.0535

=0.072Mol/L.

If this is wrong

then Simply Try The formula for Mixing of solutions

C1V1 = C2V2

0.15 x 38.5 = C2 x (15+38.5)

C2 = 0.11M/L.

7 0
3 years ago
A 13.1-g sample of CaCl2 is dissolved in 104 g of water, with both substances at 24.7°C. Calculate the final temperature of the
likoan [24]

Answer:

The final temperature of the solution is 44.8 °C

Explanation:

assuming no heat loss to the surroundings, all the heat of solution (due to the dissolving process) is absorbed by the same solution and therefore:

Q dis + Q sol = 0

Using tables , can be found that the heat of solution of CaCl2 at 25°C (≈24.7 °C)  is q dis= -83.3 KJ/mol . And the molecular weight is

M = 1*40 g/mol + 2* 35.45 g/mol = 110.9 g/mol

Q dis = q dis * n  = q dis * m/M =  -83.3 KJ/mol * 13.1 g/110.9 gr/mol = -9.84 KJ

Qdis= -9.84 KJ

Also Qsol = ms * Cs * (T - Ti)

therefore

ms * Cs * (T - Ti) + Qdis = 0

T=  Ti - Qdis * (ms * Cs )^-1   =24.7 °C - (-9.84 KJ/mol)/[(104 g + 13.1 g)* 4.18 J/g°C] *1000 J/KJ

T= 44.8 °C

7 0
4 years ago
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