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
Answer:
1.8 × 10⁻¹⁶ mol
Explanation:
(a) Calculate the solubility of the Sr₃(PO₄)₂
Let s = the solubility of Sr₃(PO₄)₂.
The equation for the equilibrium is
Sr₃(PO₄)₂(s) ⇌ 3Sr²⁺(aq) + 2PO₄³⁻(aq); Ksp = 1.0 × 10⁻³¹
1.2 + 3s 2s
![K_{sp} =\text{[Sr$^{2+}$]$^{3}$[PO$_{4}^{3-}$]$^{2}$} = (1.2 + 3s)^{3}\times (2s)^{2} = 1.0 \times 10^{-31}\\\text{Assume } 3s \ll 1.2\\1.2^{3} \times 4s^{2} = 1.0 \times 10^{-31}\\6.91s^{2} = 1.0 \times 10^{-31}\\s^{2} = \dfrac{1.0 \times 10^{-31}}{6.91} = 1.45 \times 10^{-32}\\\\s = \sqrt{ 1.45 \times 10^{-32}} = 1.20 \times 10^{-16} \text{ mol/L}\\](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BSr%24%5E%7B2%2B%7D%24%5D%24%5E%7B3%7D%24%5BPO%24_%7B4%7D%5E%7B3-%7D%24%5D%24%5E%7B2%7D%24%7D%20%3D%20%281.2%20%2B%203s%29%5E%7B3%7D%5Ctimes%20%282s%29%5E%7B2%7D%20%3D%20%201.0%20%5Ctimes%2010%5E%7B-31%7D%5C%5C%5Ctext%7BAssume%20%7D%203s%20%5Cll%201.2%5C%5C1.2%5E%7B3%7D%20%5Ctimes%204s%5E%7B2%7D%20%3D%201.0%20%5Ctimes%2010%5E%7B-31%7D%5C%5C6.91s%5E%7B2%7D%20%3D%201.0%20%5Ctimes%2010%5E%7B-31%7D%5C%5Cs%5E%7B2%7D%20%3D%20%5Cdfrac%7B1.0%20%5Ctimes%2010%5E%7B-31%7D%7D%7B6.91%7D%20%3D%201.45%20%5Ctimes%2010%5E%7B-32%7D%5C%5C%5C%5Cs%20%3D%20%5Csqrt%7B%201.45%20%5Ctimes%2010%5E%7B-32%7D%7D%20%3D%201.20%20%5Ctimes%2010%5E%7B-16%7D%20%5Ctext%7B%20mol%2FL%7D%5C%5C)
(b) Concentration of PO₄³⁻
[PO₄³⁻] = 2s = 2 × 1.20× 10⁻¹⁶ mol·L⁻¹ = 2.41× 10⁻¹⁶ mol·L⁻¹
(c) Moles of PO₄³⁻
Moles = 0.750 L × 2.41 × 10⁻¹⁶ mol·L⁻¹ = 1.8 × 10⁻¹⁶ mol
I would say the answer is C) because the thermal energy of the coffee going into the cup has to level out until the cup gets warmer and the coffee gets cooler, and they reach the same temperature.Meaning that the jug which has already been at the same temperature(we can assume based on the verbiage “the coffee was poured from the jug to the cup”) which would give the jug more thermal energy.
Answer:
The new pressure of the gas in the chamber is 1,093.75 mmHg
Explanation:
The Gay-Lussac Law is a gas law that relates pressure and temperature to constant volume. This law says that the pressure of the gas is directly proportional to its temperature.
That is, if the temperature increases, the pressure increases, while if the temperature decreases, the pressure decreases. So the Gay-Lussac law can be expressed mathematically as follows:

Having an initial and an end state of a gas, the following expression can be used:

In this case:
- P1= 1560 mmHg
- T1= 445 K
- P2=?
- T2= 312 K
Replacing:

Solving:

P2=1,093.75 mmHg
<u><em>The new pressure of the gas in the chamber is 1,093.75 mmHg</em></u>
Answer: 
Explanation:
First we have to calculate the heat absorbed by bomb calorimeter
Formula used :

= heat absorbed by calorimeter = ?
= specific heat of = 10.69 kJ/K
= final temperature = 
= initial temperature = 

As heat absorbed by calorimeter is equal to the heat released by acetylene during combustion.
Thus 1.53 gram of acetylene releases heat of combustion = 258.7kJ
So, 26.04 g/mole of acetylene releases heat of combustion 
Therefore, the heat of combustion of acetylene is, 