Answer:
Your answer would be D, Hope this helps.
<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
Answer:
To calculate the volume we must first find the number of moles
Number of moles (n ) = mass / Molar mass (M)
Since oxygen is diatomic
M of oxygen = 16 × 2 = 32g/mol
n = 16 / 32 = 0.5mol
Next we use the formula
V = n × V(dm³)
where V is the volume
V(dm³) is the volume of 1 mole of a substance at s.t.p which is
22.4dm³
Volume of oxygen gas at s.t.p is
0.5 × 22.4dm³
= 11.20dm³
Hope this helps you
1 mole of any gas under STP has volume =22.4 L
4.76 mol*22.4 L/1 mol ≈ 107 L
