<h3>
Answer:</h3>
89.88° C
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of gold cylinder as 75 g
- specific heat of gold is 0.129 J/g°C
- Initial temperature of gold cylinder is 65°C
- Mass of water is 500 g
- Initial temperature of water is 90 °C
We are required to calculate the final temperature;
- We know that Quantity of heat is given by the product of mass, specific heat capacity and change in temperature.
<h3>Step 1: Calculate the quantity of heat absorbed by the Gold cylinder</h3>
Assuming the final temperature is X° C
Then; ΔT = (X-65)°C
Therefore;
Q = 75 g × 0.129 J/g°C × (X-65)°C
= 9.675X - 628.875 Joules
<h3>Step 2: Calculate the quantity of heat released by water</h3>
Taking the final temperature as X° C
Change in temperature, ΔT = (90 - X)° C
Specific heat capacity of water is 4.184 J/g°C
Therefore;
Q = 500 g × 4.184 J/g°C × (90 - X)° C
= 188,280 -2092X joules
<h3>Step 3: Calculate the final temperature, X°C</h3>
we know that the heat gained by gold cylinder is equal to the heat released by water.
9.675X - 628.875 Joules = 188,280 -2092X joules
2101.675 X = 188908.875
X = 89.88° C
Thus, the final temperature is 89.88° C
That would be a depression..
Answer:
240.17 g Ba3(PO4)2
Explanation:
1. Determine the limiting reactant.
2H3PO4 + 3Ba(OH)2 --> Ba3(PO4)2 + 6H2O
moles H3PO4 = M x V = 3 x 0.286 = .858 moles H3PO4
moles Ba(OH)2 = M x V = 1.4 x 0.855 = 1.197 moles Ba(OH)2
ratio Ba(OH)2 : H3PO4 = 1.197: .858 = 1.39: 1
stoichiometric ratio Ba(OH)2 : H3PO4 = 3:2
Ba(OH)2is the limiting reactant
MM Ba3(PO4)2 = 601.92 g/mol
g Ba3(PO4)2 = moles Ba(OH)2 x(1 mol Ba3(PO4)2/3 moles Ba(OH)2) x (MM Ba3(PO4)2/ 1mol Ba3(PO4)2) = 1.197 x 1/3 x 601.92 = 240.17 g Ba3(PO4)2
Is this a typo if it isn’t a typo you can probably find the answer in google I hope you do!
Hope this helped :)
Answer:
mi = 31.28 g
Explanation:
According to the law of conservation of energy:
where,
mi = mass of ice = ?
L = Latent heat of fusion of ice = 334 J/g
mw = mass of water = 50 g
C = specific heat of water = 4.18 J/g.°C
ΔT = change in temperature of water = 75°C - 25°C = 50°C
Therefore,
<u>mi = 31.28 g</u>
