Answer:
Explanation:
Hello!
In this case, since the energy involved during a heating process is shown below:
Whereas the specific heat of water is 4.184 J/(g°C), we can compute the heated mass of water by the addition of 11.9 kJ (11900 J) of heat as shown below:
Thus, by plugging in, we obtain:
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Answer: 150 grams
Explanation: m = V × ρ
= 15 milliliter × 10 gram/cubic centimeter
= 15 cubic centimeter × 10 gram/cubic centimeter
= 150 gram
Answer:
0.125 moles
Explanation:
2.8 litres is equivalent to 2.8dm³
At STP,
1 mole = 22.4 dm³
x mole = 2.8 dm³
Cross multiply
22.4x = 2.8
Divide both sides by 22.4
x = 2.8/22.4
x = 0.125
Br2 experiences dipole-dipole interactions. ICl experiences dipole-dipole interactions. Br2 forms hydrogen bonds. ICl experiences induced dipole-induced dipole interactions.
It's simple, just follow my steps.
1º - in 1 L we have
of
2º - let's find the number of moles.
3º - The concentration will be
But we have this reaction
This concentration will be the concentration of
![K_{sp}=\frac{[Ba^{2+}][CO_3^{2-}]}{[BaCO_3]}](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cfrac%7B%5BBa%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D%7D%7B%5BBaCO_3%5D%7D)
considering
![[BaCO_3]=1~mol/L](https://tex.z-dn.net/?f=%5BBaCO_3%5D%3D1~mol%2FL)
![K_{sp}=[Ba^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BBa%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
and
![[Ba^{2+}]=[CO_3^{2-}]=5.07\times10^{-5}~mol/L](https://tex.z-dn.net/?f=%5BBa%5E%7B2%2B%7D%5D%3D%5BCO_3%5E%7B2-%7D%5D%3D5.07%5Ctimes10%5E%7B-5%7D~mol%2FL)
We can replace it
Therefore the
is:
