Answer:
This means that a solution must be cooled to a lower temperature than the pure solvent in order for freezing to occur.
Explanation:
Answer:
We'll have 1 mol Al2O3 and 3 moles H2
Explanation:
Step 1: data given
Numer of moles of aluminium = 2 moles
Number of moles of H2O = 6 moles
Step 2: The balanced equation
2Al + 3H2O → Al2O3 + 3H2
Step 3: Calculate the limiting reactant
For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2
Aluminium is the limiting reactant. It will completely be consumed (2 moles).
H2O is in excess. There will react 3/2 * 2 = 3 moles
There will remain 6 - 3 = 3 moles
Step 4: Calculate moles products
For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2
For 2 moles Al we'll have 2/1 = 1 mol Al2O3
For 2 moles Al We'll have 3/2 * 2 = 3 moles H2
We'll have 1 mol Al2O3 and 3 moles H2
Answer:
<h2>1.11 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula

From the question we have

We have the final answer as
<h3>1.11 g/mL</h3>
Hope this helps you
Answer:
Both
Explanation:
The combined gas law is also known as the general gas law.
From the ideal gas law we assume that n = 1;
So;
PV = nRT
and then;
=
If we cross multiply;
P₁V₁T₂ = P₂V₂T₁
So;
T₁ = 
Also;
V₂ = 
So from the choices both are correct
The question is incomplete. The complete question is :
A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)
Solutions :
If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.
Let's take :


The rate constant =
respectively.
The activation energy and the Arhenius factor is same.
So by the arhenius equation,
and 




Given,
J/mol
R = 8.314 J/mol/K





∴ 
So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.