Answer:
5
Explanation:
Firstly, we convert what we have to percentage compositions.
There are two parts in the molecule, the sulphate part and the water part.
The percentage compositions is as follows:
Sulphate- (103.74)/(103.74 + 58.55) × 100% = apprx 64%
The water part = 100 - 64 = 36%
Now, we divide the percentages by the molar masses.
For the CuSO4 molar mass is 64 + 32 + 4(16) = 160g/mol
For the H2O = 2(1) + 16 = 18g/mol
Now we divide the percentages by these masses
Sulphate = 64/160 = 0.4
Water = 36/18 = 2
The ratio is thus 0.4:2 = 1:5
Hence, there are 5 water molecules.
Answer: 1.2642*10²⁵ on both sides
Explanation:
First check how many moles are there on each side.
Since this is a balanaced equataion the number of moles on each side is the same thus the number of atoms is also same on both sides
There are 3 moles of carbon and 8 moles of hydrogen in C3H8
and 2 moles of oxygen in O2 but there 5 infront so 2*5 is 10
Number of moles on the right is 10+8+3 = 21
Now use Avogrado's constant
21 Moles* (6.02*10²³)/Mol
= 21*6.02*10²³
= 1.2642*10²⁵
The heat lost by the metal should be equal to the heat
gained by the water. We know that the heat capacity of water is simply 4.186 J
/ g °C. Therefore:
100 g * 4.186 J / g °C * (31°C – 25.1°C) = 28.2 g * Cp *
(95.2°C - 31°C)
<span>Cp = 1.36 J / g °C</span>
9.74x 2351 that's the answer
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