Answer:
The questions are incomplete
Explanation:
(b) This question is incomplete. However to calculate the number of moles of CuSO₄ present in the impure sample. The formula below can be used;
number of moles = mass of CuSO₄ present in the impure sample ÷ molar mass of CuSO₄
(c) This question is also incomplete. However, to calculate the mass percentage of CuSO₄ present in the impure sample of CuSO₄, the formula below can be used.
Mass percentage of CuSO₄ =
mass of CuSO₄ present in the impure sample/mass of impure CuSO₄ × 100
Thus, the mass of the impure sample must be measured also since the actual mass of the CuSO₄ present in the impure sample must have been measured before calculating the number of moles.
Energy required to raise the temperature from 35°C - 45 °C= 25116 J.
specific heat, the quantity of warmth required to raise the temperature of one gram of a substance by means of one Celsius degree. The units of precise warmth are generally energy or joules consistent with gram according to Celsius diploma. for instance, the unique warmth of water is 1 calorie (or 4.186 joules) according to gram in step with Celsius degree.
solving,
Sample of liquid = 400. 0 g
temperature = 30. 0 ºc
joules of energy are required to raise the temperature of the water to 45. 0 ºc
therefore rise in temperature 45 - 30 = 15°C
Specific heat capacity = 4.186 J/g m °C
In kelvin = 273 + 15 = 288
= ∴ energy required = Q = m s ( t final - t initial)
= 400*4.186 * 15
= 25116 joule
Learn more about specific heat here:-brainly.com/question/21406849
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The population would decrease, the more predators there are the more food needed for the species .
Answer:
The boiling point of a 8.5 m solution of Mg3(PO4)2 in water is<u> 394.91 K.</u>
Explanation:
The formula for molal boiling Point elevation is :

= elevation in boiling Point
= Boiling point constant( ebullioscopic constant)
m = molality of the solution
<em>i =</em> Van't Hoff Factor
Van't Hoff Factor = It takes into accounts,The abnormal values of Temperature change due to association and dissociation .
In solution Mg3(PO4)2 dissociates as follow :

Total ions after dissociation in solution :
= 3 ions of Mg + 2 ions of phosphate
Total ions = 5
<em>i =</em> Van't Hoff Factor = 5
m = 8.5 m
= 0.512 °C/m
Insert the values and calculate temperature change:



Boiling point of pure water = 100°C = 273.15 +100 = 373.15 K

= 373.15 K[/tex]
21.76 = T - 373.15
T = 373.15 + 21.76
T =394.91 K
Answer:
1. pH = 1.23.
2. 
Explanation:
Hello!
1. In this case, for the ionization of H2C2O4, we can write:

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the pKa is:

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

Which is also shown in net ionic notation.
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