5.512 litres is the volume of 15.2 grams of sulphur dioxide gas at STP.
Explanation:
Data given:
mass of sulphur dioxide = 15.2 grams
conditions is at STP whech means volume = 22.4 litres
atomic mass of sulphur dioxide = 64.06 grams/mole
Number of moles is calculated as:
number of moles = 
Putting the values in the equation:
number of moles = 
= 0.23 moles
Assuming that sulphur dioxide behaves as an ideal gas, we can calculate the volume as:
When 1 mole of sulphur dioxide occupies 22.4 litres at STP
Then 0.23 moles of sulphur dioxide occupies 22.4 x 0.23
= 5.152 litres is the volume.
Answer:
2.64%
Explanation:
mass percent = (grams of solute / grams of solution) x 100
mass percent = (2.4 / 91) × 100
mass percent = 2.64% to 3sf
Answer:
2 atoms
Explanation:
1 sodium atom + 1 chlorine atom = 2 atoms
<span>CH4 + 4 Cl2 → CCl4 + 4 HCl
(4.00 mol CH4) x (1/1) x (0.70) = 2.80 mol CCl4
(4.00 mol CH4) x (4/1) x (0.70) = 11.2 mol HCl
CCl4 + 2 HF → CCl2F2 + 2 HCl
(2.80 mol CCl4) x (2/1) x (0.70) = 3.92 mol HCl
11.2 mol + 3.92 mol = 15.1 mol HCl from both steps</span>
Answer: 1,013.32 cal × 4.18 J/cal = 4,235.68 J
Explanation:
1) Data:
Water ⇒ C = 1 cal/g°C
m = 65.8 g
Ti = 31.5°C
Tf = 36.9°C
Heat, Q = ?
2) Formula:
Q = mCΔT
3) Calculations:
Q = 65.8g × 1 cal/g°C × (46.9°C - 31.5°C) = 1,013.2 cal
4) You can convert from calories to Joules using the conversion factor:
1 cal = 4.18 J
⇒ 1,013.32 cal × 4.18 J/cal = 4,235.68 J
