Answer:
Will group 2 elements lose electrons to bond with nonmetals in group 17 in a
1:2 ratio?
Explanation:
Metals are electropositive in nature. This means that they loose electrons. Thus, metals form ionic bonds by loosing electrons to non metals.
Elements of group 2 have a valency of 2 while those of group 17 has a valency of 1 so the ratio in which group 2 elements bond with elements of group 17 is 1:2. Hence the answer.
If this is for chemistry then it's "made up of atoms."
Answer:
Option C. +150KJ
Explanation:
Data obtained from the question include:
Heat of reactant (Hr) = 200KJ
Heat of product (Hp) = 350KJ
Change in enthalphy (ΔH) =..?
The enthalphy of the reaction can be obtained as follow:
Change in enthalphy (ΔH) = Heat of reactant (Hp) – Heat of reactant (Hr)
ΔH = Hp – Hr
ΔH = 350 – 200
ΔH = +150KJ
Therefore, the enthalphy for the reaction above is +150KJ
Answer:
do you have notes for this if I saw an example I might know the answer
Answer:
83ºC
Explanation:
A bomb calorimeter is an instrument used to measure the heat that release or absorb a particular reaction.
The reaction of combustion of propane is:
C₃H₈ + 5O₂ → 3 CO₂ + 4 H₂O ΔH = -2222kJ/mol
<em>1 mole of propane release 2222kJ</em>
10.0g of propane (Molar mass: 44.1g/mol).
10.0g ₓ (1mol/ 44.1g) = <em>0.227 moles of C₃H₈</em>
If 1 mole of propane release 2222kJ, 0.227moles will release (Release because molar heat is < 0):
0.227 moles of C₃H₈ ₓ (2222kJ / mol) = 504kJ.
Our calorimeter has a constant of 8.0kJ/ºC, that means if there are released 8.0kJ, the bomb calorimeter will increase its temperature in 1ºC. As there are released 504kJ:
504kJ ₓ (1ºC / 8.0kJ) = 63ºC will increase the temperature in the bomb calorimeter.
As initial temperature was 20ºC, final temperature will be:
<h2>83ºC</h2>
