As,
Water has a pkw=14
so it can be represented as,
[H+] [OH-] = 1*10^-14
If [H+] = 3*10^-5M
[OH-] = (1*10^-14) / ( 3*10^-5)
[OH-] = 3.3*10^-9 M
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:
The last one because the first one is the most conductive so if you reverse its least conductive
Explanation:
tadaaa
Answer:

Explanation:
We must convert formula units of Zn(ClO₃)₂ to moles and then to grams of Zn(ClO₃)₂.
Step 1. Convert formula units to moles

Step 2. Convert moles to grams

Answer:
148.04 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
NO(g) + 1/2 O₂(g) → NO₂(g) ΔH°rxn = -114.14 kJ/mol
We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.
ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))
ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol
ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol
ΔH°f(NO(g)) = 148.04 kJ/mol