Hello!
First you need to calculate q
<span>delta U is change in internal energy </span>
<span>delta U = q + w </span>
<span>q is heat and w work done </span>
<span>here work was done by the system means energy leaving the system so w is negative </span>
<span>delta U = q + w </span>
<span>q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ </span>
<span>q = m x c x delta T </span>
<span>7211 J = 80.0 g x c x (225-25) °C </span>
<span>c = 0.451 J /g °C
</span>
Hope this Helps! Have A Wonderful Day! :)
The oxidation number of H is -1.
Sum of the oxidation numbers in each element =
charge of the complex
CaH₂ has 1 Ca atom and 2H atoms. The charge of
the complex is zero. Let’s say Oxidation number of H is "a".
Then,
<span> (+2)
+ 2 x a = 0 </span>
<span> +2 + 2a = 0</span>
2a = -2
a = -1
Hence, the oxidation number of Hydrogen atom in CaH₂ is -1
Moles of N2O5 = moles of NO2 * ( 2 moles of N2O5 / 4 moles of NO2
Answer:
My guess is b or c but its robably wrong
Explanation:
I just also need points sorry <3
Answer:
The choice of the answer is fourth option that is -61 degrees.
Therefore the temperature drop is -61°Centigrade.
Explanation:
Given:
The temperature in a town started out at 55 degrees
Start temperature = 55°Centigrade. (Initial temperature)
End of the Day = -6°Centigrade. (Final temperature)
To Find:
How far did the temperature drop?
Solution:
We will have,
Substituting the above values in it we get
Therefore the temperature drop is -61°Centigrade.
