Answer:
Explanation: I think its 4.91 x 10^25. Im not very sure, i just multipled 1.15 mol by the molar mass of Cl 2, which was 70.9 g. Then I multiplied that by avogadro's number. sorry if im wrong
Answer is: atoms have the greatest kinetic energy in liquid iron (2,000°C).
The average kinetic energy of molecules depends on the temperature.
At high temperature(2,000°C), molecules of iron have greater kinetic energy than molecules of iron at low temperature (1,600°C, 65°C and 25°C)..
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The answer is 19.9 grams cadmium.
Assuming there was no heat leaked from the system, the heat q lost by cadmium would be equal to the heat gained by the water:
heat lost by cadmium = heat gained by the water
-qcadmium = qwater
Since q is equal to mcΔT, we can now calculate for the mass m of the cadmium sample:
-qcadmium = qwater
-(mcadmium)(0.850J/g°C)(38.6°C-98.0°C)) = 150.0g(4.18J/g°C)(38.6°C-37.0°C)
mcadmium = 19.9 grams
Mass % of NaCl = 0.9 % (m/v)
Density of 0.9 % NaCl is about 1.0 g/mL
Volume of NaCl in mL :
![1.25 L solution * \frac{1000 mL}{1 L} * \frac{0.9 g NaCl}{100 mL solution}*\frac{1 mL}{1.0 g NaCl}](https://tex.z-dn.net/?f=%201.25%20L%20solution%20%2A%20%5Cfrac%7B1000%20mL%7D%7B1%20L%7D%20%2A%20%5Cfrac%7B0.9%20g%20NaCl%7D%7B100%20mL%20solution%7D%2A%5Cfrac%7B1%20mL%7D%7B1.0%20g%20NaCl%7D%20%20%20)
= 11.25 mL NaCl
Answer:
![pH=11.12](https://tex.z-dn.net/?f=pH%3D11.12)
Explanation:
Hello,
In this case, ammonia dissociation is:
![NH_3(aq)+H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)](https://tex.z-dn.net/?f=NH_3%28aq%29%2BH_2O%28l%29%5Crightleftharpoons%20NH_4%5E%2B%28aq%29%2BOH%5E-%28aq%29)
So the equilibrium expression:
![Kb=\frac{[NH_4^+][OH^-]}{[NH_3]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BNH_4%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BNH_3%5D%7D)
That in terms of the reaction extent and the initial concentration of ammonia is written as:
![1.8x10^{-5}=\frac{x*x}{0.10M-x}](https://tex.z-dn.net/?f=1.8x10%5E%7B-5%7D%3D%5Cfrac%7Bx%2Ax%7D%7B0.10M-x%7D)
Thus, solving by using solver or quadratic equation we find:
![x=0.00133M](https://tex.z-dn.net/?f=x%3D0.00133M)
Which actually equals the concentration of hydroxyl ion, therefore the pOH is computed:
![pOH=-log([OH^-])=-log(0.00133)=2.88](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29%3D-log%280.00133%29%3D2.88)
And the pH from the pOH is:
![pH=14-pOH=14-2.88\\\\pH=11.12](https://tex.z-dn.net/?f=pH%3D14-pOH%3D14-2.88%5C%5C%5C%5CpH%3D11.12)
Best regards.