Answer:
625.46 °C
Explanation:
We'll begin by converting 19 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
T(°C) = 19 °C
T(K) = 19 °C + 273
T(K) = 292 K
Next, we shall determine the Final temperature. This can be obtained as follow:
Initial volume (V₁) = 3.25 L
Initial temperature (T₁) = 292 K
Final volume (V₂) = 10 L
Final temperature (T₂) =?
V₁/T₁ = V₂/T₂
3.25 / 292 = 10 / T₂
Cross multiply
3.25 × T₂ = 292 × 10
3.25 × T₂ = 2920
Divide both side by 3.25
T₂ = 2920 / 3.25
T₂ = 898.46 K
Finally, we shall convert 898.46 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
T(K) = 898.46 K
T(°C) = 898.46 – 273
T(°C) = 625.46 °C
Therefore the final temperature of the gas is 625.46 °C
Answer:
Explanation: Q1 = mc(ice) ΔT (ice warms)
Q2 = ms (ice melts)
Q3 = mc((water) ΔT (water warms)
Q4 = mr (water boils)
Q5 = mc(vapour)ΔT
Answer:
it gets rust and could not run properly
Answer:
Mass = 88.12 g
Explanation:
Given data:
Mass of iron oxide = 126 g
Mass of iron formed = ?
Solution:
Chemical equation:
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Number of moles of iron oxide:
Number of moles = mass/molar mass
Number of moles = 126 g/ 159.69 g/mol
Number of moles = 0.789 mol
Now we will compare the moles of iron with iron oxide.
Fe₂O₃ : Fe
1 : 2
0.789 : 2/1×0.789 = 1.578 mol
Mass of iron:
Mass = number of moles ×molar mass
Mass = 1.578 mol × 55.84 g/mol
Mass = 88.12 g
