12 moles are in the 175L of SO2 gas at STP
An atom hopefully this helps
Answer is: the percent by mass of NaHCO₃ is 2,43%.
m(NaHCO₃) = 10 g.
V(H₂O) = 400 ml.
d(H₂O) = 1 g/ml.
m(H₂O) = V(H₂O) · d(H₂O).
m(H₂O) = 400 ml · 1 g/ml.
m(H₂O) = 400 g.
m(solution) = m(H₂O) + m(NaHCO₃).
m(solution) = 400 g + 10 g.
m(solution) = 410 g.
ω(NaHCO₃) = 10 g ÷ 410 g · 100%.
ω(NaHCO₃) = 2,43 %
The reactant is what you begin with.
The product is what you end up with (so the answer is B)
Answer:
5.06atm
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (atm)
P2 = final pressure (atm)
V1 = initial volume (Litres)
V2 = final volume (Litres)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
P1 = 1.34 atm
P2 = ?
V1 = 5.48 L
V2 = 1.32 L
T1 = 61 °C = 61 + 273 = 334K
T2 = 31 °C = 31 + 273 = 304K
Using P1V1/T1 = P2V2/T2
1.34 × 5.48/334 = P2 × 1.32/304
7.34/334 = 1.32P2/304
Cross multiply
334 × 1.32P2 = 304 × 7.34
440.88P2 = 2231.36
P2 = 2231.36/440.88
P2 = 5.06
The final pressure is 5.06atm
