Answer:
Explanation:
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In this case, since the heat involved during a heating process is computed in terms of mass, specific heat and temperature change as shown below:
Thus, since the heated mass of water was 88 g, the specific heat of water is 4.184 J/g°C and the temperature change is 6.0 °C, we can compute the heat as shown below:
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Answer:
Reaction will proceed towards forward direction.
Explanation:
Balanced reaction: 
Reaction quotient (
) for this reaction is represented as:
![Q_{c}=\frac{[COCl_{2}]}{[CO][Cl_{2}]}](https://tex.z-dn.net/?f=Q_%7Bc%7D%3D%5Cfrac%7B%5BCOCl_%7B2%7D%5D%7D%7B%5BCO%5D%5BCl_%7B2%7D%5D%7D)
Here, [CO] = [
] = 0.010 M and [
] = 0.070 M
So, 
= 700
As > 
therefore reaction will proceed towards forward direction i.e. more will be produced.
Q=mcT
125cal=(60.0g)(0.0920)(T-21)
[(125)/(60)(0.0920)]+21=T2
T2=43.6C
The thermal energy of a substance is related to the movement of the particles in the substance. This energy is defined by the temperature of the substance and whether heat is transferred or lost. This can be explained as a higher temperature causes the particles in substances to move faster and collide.
Answer:
23.5 grams of AlBr3 will be produced by 27.20 grams of NaBr
Explanation:
The balanced equation here is
6NaBr + 1AlO3 = 3Na2O + 2AlBr3
6 moles of NaBr are required to produce 2 moles of AlBr3
Mass of one mole of NaBr = 102.894 g/mol
Mass of one mole of AlBr3 = 266.69 g/mol
Mass of 6 moles of NaBr = 6*102.894 g/mol
Mass of two moles of AlBr3 = 2*266.69 g/mol
6*102.894 g NaBr produces 2*266.69 g of AlBr3
23.5 grams of AlBr3 will be produced by (6*102.894)/(2*266.69 )*23.5 = 27.20 grams of NaBr
