Answer:
Volume occupied by 55.5 moles of water gas at STP = 1240 Liters.
Explanation:
1st keep in mind that 'volume' in the context of problems like this apply only to the gas phase form of the substance. Water in solid form (ice) or liquid form (liquid water) do not apply. Volume in this case is referred to as 'molar volume' and is a gas occupying 22.4 liters at STP conditions (0°C, 1Atm).
So, if 1 molar volume of water gas (steam) occupies 22.4 Liters at STP, then 55.5 moles of water gas occupies 22.4 Liters/mole x 55.5 moles = 1243.2 Liters, or 1240 Liters (3 sig. figs.). :-)
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.
Answer : The concentration of nitrate ions is 0.450 m.
Solution : Given,
Molarity of 
= 0.225 m
The balanced ionic equation is,
From the reaction, we conclude that the 1 mole of produces 2 moles of 
ions. That means the concentration of nitrate ions is twice the value of .
then, 0.225 m of gives 2 × 0.225 of ions
Now the concentration of ions is equal to 0.450 m.
Answer:
250 mmol are left after 26.60 secs ≅ 26.6 seconds
Explanation:
zero order kinetics formula is:
[A] = [A₀] - kt
where [A] = amount left; [A₀] = amount remaining; k = rate constant; t = time
concentration [A₀] = mole/volume = 0.5 mole/2.0L = 0.25
[A] = 0.25 mole/2.0L = 0.125 M
k = 0.0047 Ms⁻¹
t = {[A]-[A₀]}/-k = (0.125 - 0.25)/(-0.0047) = 26.60 seconds = 26.6 seconds
the amount is reduced by half after 26.6 seconds
