Answer:
increased
Explanation:
I believe to be increased ,because the coal is a natural resource to be used.
i really hope it helps
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of
) =-1275.0
enthalpy of combustion of oxygen(Δ
of
) = zero
enthalpy of combustion of carbon dioxide(Δ
of
) = -393.5
enthalpy of combustion of water(Δ
of
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ
= ∈Δ
(products) - ∈Δ
(reactants)
(s) +6
(g) → 6
(g)+ 6
(l)
Δ
= [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ
= [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ
= 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ
= -2361 - 1714 - 0 + 1275
Δ
=-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Answer:
This question is incomplete
Explanation:
The question is incomplete because of the absence of the table but since the question says there are data from an investigation about a plant growth and five other plants (making six) of the same type, the best way to display this type of data for analyst is to use the grouped bar chart. <u>The grouped bar chart will display the data obtained (from an investigation on plant growth) from different students on each of the six plants (of the the same type)</u>.
Colours are usually used to identify the bars (of a group) or could be used to separate the group from other groups but in this case, colours are better used to identify the bars of a group.
Answer:
Explanation:
Hello there!
In this case, according to the Charles' law equation which help us to understand the directly proportional relationship between volume and temperature:
Thus, by solving for the final temperature, T2, and making sure we use the temperatures in Kelvin, we can calculate the final temperature as shown below:

Best regards!
Best regards!