<span>4 Al + 3 O2 → 2 Al2O3
(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2
0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.
((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =
10.1 g O2 left over</span><span>
</span>
Answer: It's equal to 10^(-2.3), or 0.00501 M, or 5.01 * 10^-3 moles/Liter
Explanation:
Well, pH = - log[H+]
Or, in words, pH is equal to -1 multiplied by the logarithm (base 10) of the hydrogen ion concentration.
So you have 2.3 = -log[H+]. We want to isolate the H+, so let's start simplifying the right hand side of the equation. First, we multiply both sides by -1.
-2.3=log[H+]
Now, the definition of a logarithm says that if the log (base 10) of [H+] is -2.3, then 10 raised to the -2.3 power is [H+]
So on each side of the equation, we raise 10 to the power of that side of the equation.
10^(-2.3) = 10^(log[H+])
and because 10^log cancels out...
10^(-2.3) = [H+]
Now we've solved for [H+], the hydrogen ion concentration!
Answer:
Explanation:
<u>1) Rate law, at a given temperature:</u>
- Since all the data are obtained at the same temperature, the equilibrium constant is the same.
- Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:
r = K [A]ᵃ [B]ᵇ
<u>2) Use the data from the table</u>
- Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s
Divide r₂ by r₁: [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0
- Use the first and second set of data to find the exponent a:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s
Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]
2ᵃ = 2 ⇒ a = 1
<u>3) Write the rate law</u>
This means, that the rate is independent of reactant B and is of first order respect reactant A.
<u>4) Use any set of data to find K</u>
With the first set of data
- r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹
Result: the rate constant is K = 0.167 s⁻¹
The weight in grams = 7.93 g
Given volume = 2.00 
Given density = 0.242 g/
We need to find the Mass(weight) in grams.
To find the weight in grams we need to keep in mind that the volume and density must use the same volume unit for cancellation. So that the volume units will cancel out, leaving only the mass units.
The unit of given volume is 
and unit of volume in density is , so first we need to change the unit of volume from 
to so that the volume units will cancel out, leaving only the mass units.
1 = 16.39 (given conversion)
units get cancel out leaving the unit.
Mass = Density X Volume.
Density = 0.242 g/ and Volume = 32.78
Mass = 7.93 grams (g)
The greenhouse effect captures the sun's energy and keeps the earth warm.
