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3 years ago

Helppppp plzzz for 30 points

Chemistry

2 answers:

Alex_Xolod [135]3 years ago

6 0

Answer:

i am not sure enough for A but i think the ans is in dilute amount

Explanation:

MrRissso [65]3 years ago

5 0

Answer with Explanation:

This is expirament based Q

1) Bring a magnet near ... the Cobalt will come out of te mixture and get attracted to magnet

2) Disolve it in a solution of ethanol. The Idoine gets dissolve and the other doesnt.

Hope im right!!

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Why do chemical equations have to be equal

Darina [25.2K]

Answer:

Every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. Therefore, there must be the same number of atoms of each element on each side of a chemical equation.

Explanation:

YW :)

5 0

3 years ago

How many joules are required for melting 12.8g of ice at 0 degrees Celsius?

tino4ka555 [31]

Answer:

4,270 Joules

Explanation:

The heat of fusion of water is 334 j/g. So, the equation would be (12.8 g)(334 j/g) which comes to 4,270 joules.

Written out that's just twelve point eight times three hundred and thirty four.

4 0

3 years ago

Can you guys help me with this one too

Mrrafil [7]

C I think not sure give it a try

4 0

3 years ago

PS.

Scrat [10]

The anwser is A thank you

6 0

3 years ago

3.10 mol of an ideal gas with CV,m=3R/2 is expanded adiabatically against a constant external pressure of 1.00 bar. The initial

Otrada [13]

Answer : The value of q, w, ΔU and ΔH for the process is, 0 J, -7.78 kJ, -7.78 kJ and -12.97 kJ respectively.

Explanation :

First we have to calculate the final temperature of the gas.

Using poission equation:

$(\frac{P_1}{P_2})^{1-\gamma}=(\frac{T_2}{T_1})^{\gamma}$

where,

$P_1$ = initial pressure = 14.7 bar

$P_2$ = final pressure = 1.00 bar

$T_1$ = initial temperature = 305 K

$T_2$ = final temperature = ?

$\gamma=\frac{R}{C_v}+1=frac{R}{(\frac{3R}{2})}+1=\frac{5}{3}=1.67$

Now put all the given values in the above expression, we get:

$(\frac{14.7}{1.00})^{1-1.67}=(\frac{T_2}{305})^{1.67}$

$T_2=103.7K$

Now we have to calculate the q, w, ΔU and ΔH for the process.

As we know that, at adiabatic process the value of q=0. So,

$dq=dU-w\\\\0=dU-w\\\\dU=w\\\\w=dU=n\times C_v\times (T_2-T_1)$

$w=dU=3.10mol\times \frac{3R}{2}\times (103.7-305)K$

$w=dU=3.10mol\times \frac{3\times 8.314J/mol.K}{2}\times (103.7-305)K$

$w=dU=-7782.278J=-7.78kJ$

Now we have to calculate the value of $C_p$

$C_p-C_v=R\\\\C_p=R+C_v\\\\C_p=R+\frac{3R}{2}\\\\C_p=\frac{5R}{2}$

Now we have to calculate the value of ΔH.

$dH=n\times C_p\times (T_2-T_1)$

$dH=3.10mol\times \frac{5R}{2}\times (103.7-305)K$

$dH=3.10mol\times \frac{5\times 8.314J/mol.K}{2}\times (103.7-305)K$

$dH=-12970.5J=-12.97kJ$

Thus, the value of q, w, ΔU and ΔH for the process is, 0 J, -7.78 kJ, -7.78 kJ and -12.97 kJ respectively.

4 0

3 years ago