Answer:
0.107 mole of SO2.
Explanation:
1 mole of a gas occupy 22.4 L at standard temperature and pressure (STP).
With the above information, we can simply calculate the number of mole of SO2 that will occupy 2.4 L at STP.
This can be obtained as follow:
22.4 L contains 1 mole of SO2.
Therefore, 2.4 L will contain = 2.4/22.4 = 0.107 mole of SO2.
Therefore, 0.107 mole of SO2 is present in 2.4 L at STP.
Answer:
1900 °C
Step-by-step explanation:
This looks like a case where we can use the <em>Combined Gas Law</em> to calculate the temperature.
p₁V₁/T₁ = p₂V₂/T₂ Multiply both sides by T₂
p₁V₁T₂/T₁ = p₂V₂ Multiply each side by T₁
p₁V₁T₂ = p₂V₂T₁ Divide each side by p₁V₁
T₂ = T₁ × p₂/p₁ × V₂/V₁
=====
Data:
We must convert the pressures to a common unit. I have chosen atmospheres.
p₁ = 675 mmHg × 1atm/760 mmHg = 0.8882 atm
V₁ = 718 mL = 0.718 L
T₁ = 48 °C = 321.15 K
p₂ = 159 kPa × 1 atm/101.325 kPa = 1.569 atm
V₂ = 2.0 L
T₂ = ?
=====
Calculation:
T₂ = 321.15 × 1.569/0.8882 × 2.0/0.718
T₂ = 321.15 × 1.766 × 2.786
T₂ = 321.15 × 1.569/0.8882 × 7.786
T₂ = 1580K
T₂ = 1580 + 273.15
T₂ = 1900 °C
<em>Note</em>: The answer can have only <em>two</em> significant figures because that is all you gave for the second volume of the gas.
at 60ºC:
106 g KNO₃ -------------- 100 g (H₂O)
? g KNO₃ ------------------ 50.0 g (H₂O)
mass KNO₃ = 50.0 * 106 / 100
mass KNO₃ = 5300 / 100
<span>= </span>53 g of KNO₃
answer (2)
<span>hope this helps!</span>
Answer: C) the rate of forward reaction is not equal to the rate of backward reaction. This should be the answer.
Explanation:
Answer:
It's the A. 2-bromobutane
Explanation:
Have a good day
