Answer:
The original molarity is 1,1030M.
Explanation:
A dilution consists of the decrease of concentration of a substance in a solution (the higher the volume of the solvent, the lower the concentration).
We use the formula for dilutions:
C1 x V1 = C2 x V2
C1= (C2 xV2)/V1
C1= ( 1400 ml x 0,286 M)/363 ml
<em>C1=1,1030 M</em>
Answer:
-238.54 kJ/mol.
Explanation:
- We need to calculate the standard enthalpy of formation of liquid methanol, CH₃OH(l) that has the equation:
C(graphite) + 2H₂(g) + ½ O₂(g) → CH3OH(l) ΔHf° = ?
?? kJ/mol.
- using the information of the three equations:
(1) C(graphite) + O₂(g) → CO₂(g), ΔHf₁° = -393.5 kJ/mol
.
(2) H2(g) + ½ O₂(g) → H₂O(l), ΔHf₂° = -285.8 kJ/mol
.
(3) CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l), ΔH₃° = -726.56 kJ/mol
.
- We rearranging and add equations (1, 2, and 3) in such a way as to end up with the needed equation:
- equation (1) be as it is:
(1) C(graphite) + O₂(g) → CO₂(g), ΔHf₁° = -393.5 kJ/mol
.
- equation (2) should be multiplied by (2) and also the value of ΔHf₂°:
(2) 2H2(g) + O₂(g) → 2H₂O(l), ΔHf₂° = 2x(-285.8 kJ/mol
).
- equation (3) should be reversed and also the value of ΔH₃° should be multiplied by (-1):
(3) CO₂(g) + 2H₂O(l) → CH₃OH(l) + 3/2 O₂(g), ΔH₃° = 726.56 kJ/mol
.
- By summing the modified equations, we can get the needed equation and so:
The standard enthalpy of formation of liquid methanol, CH₃OH(l) = ΔHf₁° + 2(ΔHf₂°) - ΔH₃° = (-393.5 kJ/mol
) + 2(-285.8 kJ/mol
) - (- 726.56 kJ/mol) = -238.54 kJ/mol.
Answer:
02 is a product
Explanation:
it doesnt take part in chemical reaction so its not reactant.02 doesnt alter rate of chemical reaction so it cannot be catalyst. it is gas it is not solid so it is a product.
<span>Answer:
Benzylic radical . First Br radicals are produced which strip off a H- from the methyl group. The benzylic radical the reacts with Br2 to form benzylic bromide and another Br radical.</span>