Answer:
1.27 moles (3 s.f.)
Explanation:
Mole ratio of water: C6H14
= 14:2
= 7:1
This means that to produce 7 moles of water, 1 mole of C6H14 is needed. Or 1/7 mole of C6H14 is needed to produce 1 mole of water. So if you need 8.86 moles of water, 8.86(1/7) moles of C6H14 is needed.
For 8.86mol of water, moles of C6H14 needed
= 8.86/7
= 1.27 moles (3 s.f.)
Answer:
The bolts will run out first, so these will limit how much product can be formed.
(option C is correct)
Explanation:
Step 1: Data given
Number of bolts = 5
Number of nuts = 12
ratio N:B = 2:1
Step 2
Since the ratio N:B is 2:1 we have 2 nuts for each bolt
When we have 2 bolts we have 4 nuts
When we have 3 bolts we have 6 nuts
When we have 4 bolts we have 8 nuts
When we have 5 bolts we have 10 nuts
We don't have more bolts, but there are still 2 nuts left. This means the bolts will run out first, so these will limit how much product can be formed.
(option C is correct)
Answer:
64.0 g/mol.
Explanation:
- Thomas Graham found that, at a constant temperature and pressure the rates of effusion of various gases are inversely proportional to the square root of their masses.
<em>∨ ∝ 1/√M.</em>
where, ∨ is the rate of diffusion of the gas.
M is the molar mass of the gas.
<em>∨₁/∨₂ = √(M₂/M₁)</em>
∨₁ is the rate of effusion of the unknown gas.
∨₂ is the rate of effusion of He gas.
M₁ is the molar mass of the unknown gas.
M₂ is the molar mass of He gas (M₂ = 4.0 g/mol).
<em>∨₁/∨₂ = 0.25.</em>
∵ ∨₁/∨₂ = √(M₂/M₁)
∴ (0.25) =√(4.0 g/mol)/(M₁)
<u><em>By squaring the both sides:</em></u>
∴ (0.25)² = (4.0 g/mol)/(M₁)
∴ M₁ = (4.0 g/mol)/(0.25)² = 64.0 g/mol.
Answer:
Explanation:
<em>Waves are actually energy passing through the water, causing it to move in a circular motion. ... This phenomenon is a result of the wave's orbital motion being disturbed by the seafloor.</em>
<em />
<em>The direction a wave propagates is perpendicular to the direction it oscillates for transverse waves. A wave does not move mass in the direction of propagation; it transfers energy.</em>
Answer:
Explanation:
Hello,
In this case, the combustion of methane is shown below:
And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:
Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:
Best regards.
