<u>Answer:</u> The final temperature of water is 32.3°C
<u>Explanation:</u>
When two solutions are mixed, the amount of heat released by solution 1 (liquid water) will be equal to the amount of heat absorbed by solution 2 (liquid water)
The equation used to calculate heat released or absorbed follows:
![m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
......(1)
where,
q = heat absorbed or released
= mass of solution 1 (liquid water) = 50.0 g
= mass of solution 2 (liquid water) = 29.0 g
= final temperature = ?
= initial temperature of solution 1 = 25°C = [273 + 25] = 298 K
= initial temperature of solution 2 = 45°C = [273 + 45] = 318 K
c = specific heat of water= 4.18 J/g.K
Putting values in equation 1, we get:
![50.0\times 4.18\times (T_{final}-298)=-[29.0\times 4.18\times (T_{final}-318)]\\\\T_{final}=305.3K](https://tex.z-dn.net/?f=50.0%5Ctimes%204.18%5Ctimes%20%28T_%7Bfinal%7D-298%29%3D-%5B29.0%5Ctimes%204.18%5Ctimes%20%28T_%7Bfinal%7D-318%29%5D%5C%5C%5C%5CT_%7Bfinal%7D%3D305.3K)
Converting this into degree Celsius, we use the conversion factor:
Hence, the final temperature of water is 32.3°C
First, we need to get the molar mass of:
KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol
KCl =39.1 + 35.5 = 74.6 g/mol
O2 = 16*2 = 32 g/mol
From the given equation we can see that:
every 2 moles of KClO3 gives 3 moles of O2
when mass = moles * molar mass
∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g
and the mass of O2 then = 3 mol * 32g/mol = 96 g
so, 245.2 g of KClO3 gives 96 g of O2
A) 2.72 g of KClO3:
when 245.2 KClO3 gives → 96 g O2
2.72 g KClO3 gives → X
X = 2.72 g KClO3 * 96 g O2/245.2 KClO3
= 1.06 g of O2
B) 0.361 g KClO3:
when 245.2 g KClO3 gives → 96 g O2
0.361 g KClO3 gives → X
∴ X = 0.361g KClO3 * 96 g / 245.2 g
= 0.141 g of O2
C) 83.6 Kg KClO3:
when 245.2 g KClO3 gives → 96 g O2
83.6 Kg KClO3 gives → X
∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3
= 32.7 Kg of O2
D) 22.4 mg of KClO3:
when 245.2 g KClO3 gives → 96 g O2
22.4 mg KClO3 gives → X
∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3
= 8.8 mg of O2
(NH4)3PO4 :
N = 14 * 3 = 42
H = 1 * 12 = 12
P = 31 * 1 = 31
O = 16 * 4 = 64
-------------------------
42+12+31+64 = 149 g / mol
Hope this helps!.
Answer:
5.79 × 10^23 Oxygen atoms
Explanation:
Number of Oxygen atom in the compound = 4×3 = 12
Molar mass of Al2(SO4)3 = 342 g/mol.
No of mole = mass/molar mass = 2.74/342 = 8.01×10^-03 mole
2.74g of Al2(SO4)3 × 1 mole of Al2 (SO4)3 / 342g of Al2 (SO4)3 * 12 mole of Oxygen/ 1mole of Al(SO4)3 * 6.02×10^23 Oxygen atom/ 1 mole of Oxygen
= 5.79×10^23 Oxygen atoms
