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
Answer:
0.6258 g
Explanation:
To determine the number grams of aluminum in the above reaction;
- determine the number of moles of HCl
- determine the mole ratio,
- use the mole ratio to calculate the number of moles of aluminum.
- use RFM of Aluminum to determine the grams required.
<u>Moles </u><u>of </u><u>HCl</u>
35 mL of 2.0 M HCl
2 moles of HCl is contained in 1000 mL
x moles of HCl is contained in 35 mL
We have 0.07 moles of HCl.
<u>Mole </u><u>ratio</u>
6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)
Hence mole ratio = 6 : 2 (HCl : Al
- but moles of HCl is 0.07, therefore the moles of Al;
Therefore we have 0.0233333 moles of aluminum.
<u>Grams of </u><u>Aluminum</u>
We use the formula;
The RFM (Relative formula mass) of aluminum is 26.982g/mol.
Substitute values into the formula;
The number of grams of aluminum required to react with HCl is 0.6258 g.
Answer: acid dissociation constant Ka= 2.00×10^-7
Explanation:
For the reaction
HA + H20. ----> H3O+ A-
Initially: C. 0. 0
After : C-Cx. Cx. Cx
Ka= [H3O+][A-]/[HA]
Ka= Cx × Cx/C-Cx
Ka= C²X²/C(1-x)
Ka= Cx²/1-x
Where x is degree of dissociation = 0.1% = 0.001 and c is the concentration =0.2
Ka= 0.2(0.001²)/(1-0.001)
Ka= 2.00×10^-7
Therefore the dissociation constant is
2.00×10^-7
Covalent bonds form when electrons are shared between two nonmetals.
The answer is D only metals are shiny and highly malleable
