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Answer: b) 0.250 mol</h3>
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Work Shown:
Using the periodic table, we see that
- 1 mole of carbon = 12 grams
- 1 mole of oxygen = 16 grams
These are approximations and these values are often found underneath the atomic symbol. For example, the atomic weight listed under carbon is roughly 12.011 grams. I'm rounding to 2 sig figs in those numbers listed above.
So 1 mole of CO2 is approximately 12+2*16 = 44 grams. The 2 is there since we have 2 oxygens attached to the carbon atom.
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Since 1 mole of CO2 is 44 grams, we can use that to convert from grams to moles.
11.0 grams of CO2 = (11.0 grams)*(1 mol/44 g) = (11.0/44) mol = 0.250 mol of CO2
In short,
11.0 grams of CO2 = 0.250 mol of CO2
This is approximate.
We don't need to use any of the information in the table.
Answer:
A. How the concentration of the reactants affects the rate of a reaction
Explanation:
Let's consider a generic reaction.
A + B ⇒ Products
The generic rate law is:
rate = k × [A]ᵃ × [B]ᵇ
where,
- rate: rate of the reaction
- [A] and [B]: molar concentrations of the reactants
As we can see, the rate law shows how the concentration of the reactants affects the rate of a reaction.
The answer is liter hope this helped :)
Answer:
They experience the same pressure
Explanation:
To answer this question, we recall Pascal's, Law Pascal's law states that an increase in pressure at a point in a confined cylinder containing a fluid, there is also an equal increase at all other points in that cylinder.
According to Pascal's law the pressure if the pressure expereienced by the larger diameter piston increases, the pressure experienced by the smaller diameter piston also increases by the same amount
However considering that pressure = Force/area F1/A1 =F2/A2
thus where A1 = πD²÷4 and A2 = πD²÷ 16 we have
we have F1×4/πD² = F2×16/πD² or F1 = 4× F2
They experience the same pressure but the larger cylinder delivers four times the force transmitted from he outside to the smaller cylinder
Answer:
Explanation:
Hi! :) You didn't post the statements, but the answer should be something about conduction. Hope this helps!
