Answer is: formula of hydrate is CoCl₂· 6H₂O -c<span>obalt(II) chloride hexahydrate
</span>m(CoCl₂· xH₂O) = 1,62 g.
m(CoCl₂) = 0,88 g.
n(CoCl₂) = m(CoCl₂) ÷ M(CoCl₂)
n(CoCl₂) = 0,88 g ÷ 130 g/mol
n(CoCl₂) = 0,0068 mol.
m(H₂O) = 1,62 g - 0,88 g.
m(H₂O) = 0,74 g.
n(H₂O) = m(H₂O) ÷ m(H₂O)
n(H₂O) = 0,74 g ÷ 18 g/mol
n(H₂O) = 0,041 mol.
n(CoCl₂) : n(H₂O) = 0,0068 mol : 0,041 mol.
n(CoCl₂) : n(H₂O) = 1 : 6.
Hope this helps!! You were correct for Question 9 but incorrect for Question 10. Yes, the value comes out to be 62.4 kJ but watch out for your signs! It is definitely a positive number. Therefore, you have to consider your other choices. When I converted the 62.4 kJ to J, I found an option that matches my answer.
Answer:
Trimethylacetaldehyde
Explanation:
For the <u>unknow compound</u> we have a molar mass of 86 g/mol. We have an even value for the mass, so the compound does <u>not have nitrogen</u> and we can have several posibilities:
A) 
B) 
C) 
D) 
If we check the IR info a signal in 1730 cm-1 appears, this indicates that we have an <u>oxo group</u> (C=O). So, the D option can be discarded. The groups that can have the oxo group are: Carboxylic acids, <u>Ketones and aldehydes</u>.
We don have a signal in 3000 cm-1, so the carboxylic acid can be discarded. Now, is we check the info for the 1H NMR we only have 2 signals. If we only have 2 we will have a very<u> symmetric compound</u>.
By trial an error the find the compound <u>Trimethylacetaldehyde</u> (Figure 1).
Answer:
- <u><em>Substance A.</em></u>
Explanation:
pH is a measure of the concentration of H₃O⁺ ions in a solution.
It is defined as:
![pH=\log \dfrac{1}{[H_3O^+]}=-\log [H_3O^+]](https://tex.z-dn.net/?f=pH%3D%5Clog%20%5Cdfrac%7B1%7D%7B%5BH_3O%5E%2B%5D%7D%3D-%5Clog%20%5BH_3O%5E%2B%5D)
For the given substances
Sustance [H₃O⁺] pH = - log [H₃O⁺]
M
A 4×10⁻⁶ -log (4×10⁻⁶) = 5.4 ← answer
B 6×10⁻⁷ -log (6×10⁻⁷) = 6.2
C error - - - - - -
D 1.3×10⁻¹ -log (1.3×10⁻¹) = 0.89
Thus, the unknown substace most likely is A.