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Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol
Explanation:
The given balanced chemical reaction is,

First we have to calculate the enthalpy of reaction
.

![\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%2Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%5D-%5Bn_%7BC_8H_%7B18%7D%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
We are given:

Putting values in above equation, we get:
![-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]](https://tex.z-dn.net/?f=-1.0940%5Ctimes%2010%5E4%3D%5B%2816%5Ctimes%20-393.5%29%2B%2818%5Ctimes%20-285.8%29%5D-%5B%2825%5Ctimes%200%29%2B%282%5Ctimes%20%5CDelat%20H_f%7BC_8H_%7B18%7D%28l%29%7D%5D)

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol
Answer
5
Explanation:
We can go about this using the percentage compositions.
First, we calculate the percentage composition of the copper sulphate. This is obtainable by using the mass.
0.96/1.5 * 100 = 64%
Hence the percentage by mass of the water present is 36%
The molar mass of the anhydrous sulphate is 64 + 32 +4(16) = 160g/mol
The molar mass of the water is 2(1) + 16 = 18g/mol
Not forgetting that it is in multiples of x, the total molar mass of the water is 18x moles
The total mass of the copper sulphate hydrate is 160+ 18x
Now how do we get x? Like it is said earlier, the percentage composition is constant.
Hence, 64/100 * (160 + 18x) = 160
16000 = 64(160 + 18x)
16000 = 10,240 + 1152x
16,000 - 10,240 = 1152x
1152x = 5760
x = 5760/1152
x = 5
Answer:
kindly check the EXPLANATION SECTION
Explanation:
In order to be able to answer this question one has to consider the neutron proton ratio. Considering this ratio will allow us to determine the stability of a nuclei. The most important rule that helps us in determination of stability is that when the Neutron- Proton ratio of any nuclei ranges from to 1 to 1.5, then we say the nuclei is STABLE.
Also, we need to understand that when the Neutron- Proton ratio is LESS THAN 1 or GREATER THYAN 1.5, then we say the nuclei is UNSTABLE.
So, let us check which is stable and which is unstable:
a. 4 protons and 5 neutrons = Neutron- proton ratio = N/P = 5/4= stable.
b. 7 protons and 7 neutrons = Neutron- proton ratio = N/P = 7/7= 1 = stable.
c. 2 protons and 3 neutrons = Neutron- proton ratio = N/P = 3/5 =0.6 =unstable.
d. 3 protons and 0 neutrons = Neutron- proton ratio = N/P = 0/3= 0= unstable.
e. 6 protons and 5 neutrons = Neutron- proton ratio = N/P = 5/6= 0.83 = unstable.
f. 9 protons and 9 neutrons = Neutron- proton ratio = N/P = 9/9 = 1 = stable.
g. 8 protons and 7 neutrons = Neutron- proton ratio = N/P = 7/8 =0.875 = unstable.
h. 1 proton and 0 neutrons = Neutron- proton ratio = N/P = 0/1 =0 = unstable