Answer:
0 g.
Explanation:
Hello,
In this case, since the reaction between methane and oxygen is:
If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:
Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.
Regards.
Answer:
35.06 g NaCl
Explanation:
mol = 0.5 L * 1.2 M
Na mass = 22.99 g
Cl mass = 35.45 g
0.6 mol * (22.99 g + 35.45 g)/1 mol = 35.06 g NaCl
Answer:
V₂ = 111.3 mL
Explanation:
Given data:
Initial volume of gas = 50.0 mL
Initial temperature = standard = 273.15 K
Final volume = ?
Final temperature = 335 °C (335+273.15 = 608.15 K)
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 50.0 mL ×608.15 K / 273.15 k
V₂ = 30407.5 mL.K / 273.15 K
V₂ = 111.3 mL
C both are very reactive i guess
Answer:
the product of the equation is 1.0 x 10^-1
trust me, i'm a major in Biochem :)
