Hey there!
Cu(CN)₂
Find the molar mass.
Cu: 1 x 63.546 = 63.546
C: 2 x 12.01 = 24.02
N: 2 x 14.07 = 28.14
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115.706 grams
The mass of one mole of Cu(CN)₂ is 115.706 grams.
We have 4 moles.
115.706 x 4 = 463
4.00 moles of Cu(CN)₂ has a mass of 463 grams.
Hope this helps!
I think Intramolecular forces are being weakened
Answer:the pH is 12
Explanation:
First We need to understand the structure of trimethylamine
Due to the grades of the bond in the nitrogen with a hybridization sp3 is 108° approximately, then is generated a dipole magnetic at the upper side of the nitrogen, this dipole magnetic going to attract a hydrogen molecule of the water making the water more alkaline
C3H9N+ H2O --> C3H9NH + OH-
![k=\frac{[C3H9NH]*[OH-]}{[C3H9N]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5BC3H9NH%5D%2A%5BOH-%5D%7D%7B%5BC3H9N%5D%7D)
Then:
The concentration of the trimethylamine is 0.3 and the concentration of the ion C3H9NH is equal to the OH- relying on the stoichiometric equation. We could find the concentration of the OH- ion with the square root of the multiplication between k and the concentration of trimethylamine
[OH-]=
[OH-]=0.01
pH=14-(-log[OH-])
pH=12
The temperature that must be to freeze the solution would be -21.1 ° C.
<h3>How to calculate the freezing temperature of this solution?</h3>
To calculate the freezing temperature we must take into account the following information.
- Solution with a salt concentration of 10% is frozen at -6°C
- Solution with a salt concentration of 20% is frozen at -16°C
- Solution with a higher concentration is frozen at -21.1°C
According to the above, it can be inferred that the puddle has a 50% concentration of salt because they had 12 kg of water and 6 kg of salt.
So the lowest freezing temperature would be 21.1°C because the puddle is 50% concentrated.
Note: This question is incomplete because there is some missing information. Here is the missing information:
- A 10% salt solution freezes at about 20°F (-6°C), and a 20% solution freezes at 2°F (-16°C).
- The lowest freezing point obtainable for salt solutions is −21.1 °C
Learn more about freezing in: brainly.com/question/14131507
Answer is: 1160 J of heat Is required to increase the temperature.
m(Fe) = 100 g.
∆T = 40,2 - 15 = 25,2°C.
C(Fe) = 0,46 J/g•°C.
Q = m(Fe) • C • ∆T.
Q = 100 g • 0,46 J/g•°C • 25,2°C
Q = 1160 J.
C - specific heat.
