Answer:
We have to fill the acetylene tank with a pressure of 60.67 atm to ensure that you run out of each gas at the same time
Explanation:
Step 1: Data given
Volume of tank 1 = 6.50 L = oxygen
Volume of tank 2 = 4.50 L = acetylene
Pressure in the oxygen tank = 105 atm
Step 2: The balanced equation
2 C2H2 + 5 O2 → 4 CO2 + 2 H2O
For 2 mol C2H2 we need 5 moles O2 to produce 4 moles CO2 and 2 moles H2O
Step 3: Calculate the pressure of the acetylene tank
For 2 mol C2H2 we need 5 moles O2 to produce 4 moles CO2 and 2 moles H2O
This means for each 5 moles O2 we have 2 moles acetylene
Pressure (105 atm) * (2/5) * (6.50 / 4.50) = 60.67 atm
We have to fill the acetylene tank with a pressure of 60.67 atm to ensure that you run out of each gas at the same time
Answer:
conclusion
Explanation:
it can't be a hypothesis since tests are carried out to verify so it is not a theory
an introduction to an experiment only gives the basis of what we are investigating therefore nothing has been proven and the question is still unanswered
Mass of the water : 2.23 g
<h3>Furter explanation</h3>
Heat
Q = m.c.Δt
m= mass, g
c = heat capacity, for water : 4.18 J/g° C.
ΔT = temperature
Q= 140 J
Δt = 75 - 60 = 15
mass of the water :
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 %
