Answer:
<h2>
The answer is C </h2>
Explanation:
when a car begins to slow down the speed that it was going will decrease. Which means that it is an example of deceleration.
Answer:
The answer to your question is the letter C) 5648 kJ/mol
Explanation:
Data
C₁₂H₂₂O₁₁ + 12 O₂ ⇒ 12 CO₂ + 11 H₂O
H° C₁₂H₂₂O₁₁ = -2221.8 kJ/mol
H° O₂ = 0 kJ / mol
H° CO₂ = -393.5 kJ/mol
H° H₂O = -285.8 kJ/mol
Formula
ΔH° = ∑H° products - ∑H° reactants
Substitution
ΔH° = 12(-393.5) + 11(-285.8) - (-2221.8) - (0)
ΔH° = -4722 - 3143.8 + 2221.8
Result
ΔH° = -5644 kJ/mol
In warmer weather gases tend to expand and take up more room, thus increasing pressure. but in colder weather they will condense or contract and take up less space, therefore lowering the pressure of the tire in this situation.
Answer:
Q = 30284.88 j
Explanation:
Given data:
Mass of ethanol = 257 g
Cp = 2.4 j/g.°C
Chnage in temperature = ΔT = 49.1°C
Heat required = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Now we will put the values in formula.
Q = 257 g× 2.4 j/g.°C × 49.1 °C
Q = 30284.88 j
An x would represent the gained electrons
A . Would represent the valence electrons
You would just draw [ ] around the diagram
And the charge should be placed outside the brackets
