Answer:
<em>Frame of Reference</em>
The fixed background surrounding an object, assumed to be at rest and in a fixed position.
Answer:
The charge of the purple circles should be positive because they represent the nuclei.
Explanation:
Answer:
27 liters of hydrogen gas will be formed
Explanation:
Step 1: Data given
Number of moles C = 1.03 moles
Pressure H2 = 1.0 atm
Temperature = 319 K
Step 2: The balanced equation
C +H20 → CO + H2
Step 3: Calculate moles H2
For 1 mol C we need 1 mol H2O to produce 1 mol CO an 1 mol H2
For 1.03 moles C we'll have 1.03 moles H2
Step 4: Calculate volume H2
p*V = n*R*T
⇒with p = the pressure of the H2 gas = 1.0 atm
⇒with V = the volume of H2 gas = TO BE DETERMINED
⇒with n = the number of moles H2 gas = 1.03 moles
⇒with R = the gas constant = 0.08206 L*Atm/mol*K
⇒with T = the temperature = 319 K
V = (n*R*T)/p
V = (1.03 * 0.08206 *319) / 1
V = 27 L
27 liters of hydrogen gas will be formed
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
