You are measuring an Intensive property
You are left with a chloride ion, so you started off with a chlorine atom.
Cl + e- → Cl-
The dilution formula can be used to find the volume needed
c1v1 = c2v2
Where c1 is concentration and v1 is volume of the concentrated solution
And c2 is concentration and v2 is volume of the diluted solution to be prepared
c1 - 0.33 M
c2 - 0.025 M
v2 - 25 mL
Substituting these values in the equation
0.33 M x v1 = 0.025 M x 25 mL
v1 = 1.89 mL
Therefore 1.89 mL of the 0.33 M solution needs to be diluted up to 25 mL to make a 0.025 M solution
Answer:
n=1 holds two electrons and n=2 holds eight electrons.
Explanation:
Hello
In this case, since the atomic number of aluminum is 13, its electron configuration is:
In such a way, we can see that the level n=1 is filled with two electrons since the subshell s is able to hold two electrons and the level n=2 is also filled but with eight electrons as s holds two whereas p holds 6. Moreover, n=3 is holding three electrons.
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Answer:
378mL
Explanation:
The following data were obtained from the question:
Pressure (P) = 99.19 kPa
Temperature (T) = 28°C
Number of mole (n) = 0.015 mole
Volume (V) =...?
Next, we shall convert the pressure and temperature to appropriate units. This is illustrated below:
For Pressure:
101.325 KPa = 1 atm
Therefore, 99.19 kPa = 99.19/101.325 = 0.98 atm
For Temperature:
T(K) = T(°C) + 273
T(°C) = 28°C
T(K) = 28°C + 273 = 301K.
Next we shall determine the volume of N2. The volume of N2 can be obtained by using the ideal gas equation as shown below:
PV = nRT
Pressure (P) = 0.98 atm
Temperature (T) = 301K
Number of mole (n) = 0.015 mole
Gas constant (R) = 0.0821atm.L/Kmol.
Volume (V) =...?
0.98 x V = 0.015 x 0.0821 x 301
Divide both side by 0.98
V = (0.015 x 0.0821 x 301) /0.98
V = 0.378 L
Finally, we shall convert 0.378 L to millilitres (mL). This is illustrated below:
1L = 1000mL
Therefore, 0.378L = 0.378 x 1000 = 378mL
Therefore, the volume of N2 collected is 378mL
