Weak acids only partially ionize. so the retain part of the original molecule. C
Answer:
1610.7 g is the weigh for 4.64×10²⁴ atoms of Bi
Explanation:
Let's do the required conversions:
1 mol of atoms has 6.02×10²³ atoms
Bi → 1 mol of bismuth weighs 208.98 grams
Let's do the rules of three:
6.02×10²³ atoms are the amount of 1 mol of Bi
4.64×10²⁴ atoms are contained in (4.64×10²⁴ . 1) /6.02×10²³ = 7.71 moles
1 mol of Bi weighs 208.98 g
7.71 moles of Bi must weigh (7.71 . 208.98 ) /1 = 1610.7 g
Answer:
0.574moles
Explanation:
Using the general gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (Litres)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (Kelvin)
According to the information provided in the question;
- Volume (V) = 12400mL = 12400/1000 = 12.4L
- Pressure (P) = 890mmHg = 890/760 = 1.17atm
- Temperature (T) = 35°C = 35 + 273 = 308K
Hence, using PV = nRT
n = PV/RT
n = 1.17 × 12.4 ÷ 0.0821 × 308
n = 14.508 ÷ 25.287
n = 0.574moles
Therefore, the number of moles of argon gas in the cylinder is 0.574moles
Possibly decomposition but not sure
Answer:
0.0583g
Explanation:
The equation of the reaction is;
2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)
From the question, number of moles of HNO3 reacted= concentration × volume
Concentration of HNO3= 0.100 M
Volume of HNO3 = 20.00mL
Number of moles of HNO3= 0.100 × 20/1000
Number of moles of HNO3 = 2×10^-3 moles
From the reaction equation;
2 moles of HNO3 reacts with 1 mole of Mg(OH)2
2×10^-3 moles reacts with 2×10^-3 moles ×1/2 = 1 ×10^-3 moles of Mg(OH)2
But
n= m/M
Where;
n= number of moles of Mg(OH)2
m= mass of Mg(OH)2
M= molar mass of Mg(OH)2
m= n×M
m= 1×10^-3 moles × 58.3 gmol-1
m = 0.0583g
