Answer:
V = 4/3 * 3.1416 * (37x10-10)3
V = 2.12x10-25 cm3
d = m/V
d = 1.67x10-24 / 2.12x10-25 = 7.87 g/cm3
The difference in temperature, let's convert F to ºC:
ºC = -80-32/1.8 = -62.22 ºC
dT = -92.6 + 62.2 = -30.4 ºC
Answer:
sublimation
Explanation:
immediately into a gas when thermal energy is added to it. The process is called sublimation. Sublimation is the change of state from a solid to a gas without going through the liquid state.
I believe you have to label out the positive metal ion and the delocalized electrons. They're the 2 things that makes up a metal structure.
In the diagram, the circles with the + symbol are the positive metal ions, since + represents positive. And the remaining - circles are the delocalized electrons, as electrons are negative.
And for how a metal conducts electricity, since they're delocalized mobile electrons present in any metal structures, they're able to move away from the metal to the positive side of the battery and more electrons can replace their place flowing from the negative side.
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol
15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol
We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M
Concentration of NO3⁻ is 0.667 M.
Answer:
Explanation:
Hello,
In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:
![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 concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:
![pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1](https://tex.z-dn.net/?f=pH%3D-log%28Ka%29%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5CpH%3D-log%286.3x10%5E%7B-5%7D%29%2Blog%28%5Cfrac%7B0.33M%7D%7B0.42M%7D%20%29%5C%5C%5C%5CpH%3D4.1)
Regards.
