Answer:
-10778.95 J heat must be removed in order to form the ice at 15 °C.
Explanation:
Given data:
mass of steam = 25 g
Initial temperature = 118 °C
Final temperature = 15 °C
Heat released = ?
Solution:
Formula:
q = m . c . ΔT
we know that specific heat of water is 4.186 J/g.°C
ΔT = final temperature - initial temperature
ΔT = 15 °C - 118 °C
ΔT = -103 °C
now we will put the values in formula
q = m . c . ΔT
q = 25 g × 4.186 J/g.°C × -103 °C
q = -10778.95 J
so, -10778.95 J heat must be removed in order to form the ice at 15 °C.
Answer:
Equilibrium constant for
is 0.5
Equilibrium constant for decomposition of
is 
Explanation:
dissociates as follows:

initial 0.72 mol 0 0
at eq. 0.72 - 0.40 0.40 0.40
Expression for the equilibrium constant is as follows:
![k=\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
Substitute the values in the above formula to calculate equilibrium constant as follows:
![k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5B0.40%2F1%5D%5B0.40%2F1%5D%7D%7B0.32%2F1%7D%20%5C%5C%3D%5Cfrac%7B0.40%20%5Ctimes%200.40%7D%7B0.32%7D%20%5C%5C%3D0.5)
Therefore, equilibrium constant for
is 0.5
Now calculate the equilibrium constant for decomposition of 
It is given that
is decomposed.
decomposes as follows:

initial 1.0 M 0 0
at eq. concentration of
is:
![[NO_2]_{eq}=1-(0.000066) = 0.999934\ M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D1-%280.000066%29%20%3D%200.999934%5C%20M)
![[NO]_{eq}=6.6 \times 10^{-5}\ M](https://tex.z-dn.net/?f=%5BNO%5D_%7Beq%7D%3D6.6%20%5Ctimes%2010%5E%7B-5%7D%5C%20M)
Expression for equilibrium constant is as follows:
![K=\frac{[NO]^2[O_2]}{[NO_2]^2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO%5D%5E2%5BO_2%5D%7D%7B%5BNO_2%5D%5E2%7D)
Substitute the values in the above expression
![K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5B6.6%5Ctimes%2010%5E%7B-5%7D%5D%5E2%5B3.3%20%5Ctimes%2010%5E%7B-5%7D%5D%7D%7B%5B0.999934%5D%5E2%7D%20%5C%5C%3D1.79%5Ctimes%2010%5E%7B-14%7D)
Equilibrium constant for decomposition of
is 
<u>Answer:</u> The mass percent of hydrogen in methyl acetate is 8 %
<u>Explanation:</u>
The given chemical formula of methyl acetate is 
To calculate the mass percentage of hydrogen in methyl acetate, we use the equation:

Mass of hydrogen = (6 × 1) = 6 g
Mass of methyl acetate = [(3 × 12) + (6 × 1) + (2 × 16)] = 74 g
Putting values in above equation, we get:

Hence, the mass percent of hydrogen in methyl acetate is 8 %
Barfoed's test is a concoction test utilized for identifying the nearness of monosaccharides. It depends on the diminishment of copper(II) acetic acid derivation to copper(I) oxide (Cu2O), which frames a block red hasten.
Barfoed's reagent comprises of a 0.33 molar arrangement of unbiased copper acetic acid derivation in 1% acidic corrosive arrangement. The reagent does not keep well and it is, thusly, fitting to make it up when it is really required. May store uncertainly as per a few MSDS's.
Answer:
x means unknown it is an unknown value.
For example if you have 2 x you have 2 u know values.
Explanation:
If you want us to explain it further please provide a picture.