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Answer:
(a). 132 × 10^-9 s = 132 nanoseconds.
(b)..176.5 pico-seconds.
Explanation:
(a). At one torr, the first thing to do is to find the speed and that can be done by using the formula below;
Speed = [ (8 × R × T)/ Mm × π]^1/2.
Where Mm = molar mass, T = temperature and R = gas constant.
Speed= [ ( 8 × 8.314 × 300)/ 131.293 × π × 10^-3)^1/2. = 220m/s.
The next thing to do now is to calculate for the degree of collision which can be calculated by using the formula below;
Degree of collision = √2 × π × speed × d^2 × pressure/ K × T.
Note that pressure = 1 torr = 133.32 N/m^2 and d = collision diameter.
Degree of collision = √2 × π × 220 × (4.9 × 10^-10)^2 × 133.32/ 1.38 × 10^-23 × 300.
Degree of collision = 7.55 × 10^6 s^-1.
Thus, 1/ 7.55 × 10^6. = 132 × 10^-9 s = 132 nanoseconds.
(b). At one bar;
1/10^5 × 10^3 × 56.65 = 1.765 × 10^-10 = 176.5 pico-seconds.
Answer:
Mass of Ca(OH)₂ required = 0.09 g
Explanation:
Given data:
Volume of HNO₃ = 25 mL (25/1000 = 0.025 L)
Molarity of HNO₃ = 0.100 M
Mass of Ca(OH)₂ required = ?
Solution:
Chemical equation;
Ca(OH)₂ + 2HNO₃ → Ca(NO)₃ + 2H₂O
Number of moles of HNO₃:
Molarity = number of moles / volume in L
0.100 M = number of moles / 0.025 L
Number of moles = 0.100 M ×0.025 L
Number of moles = 0.0025 mol
Now we will compare the moles of Ca(OH)₂ with HNO₃ from balance chemical equation.
HNO₃ : Ca(OH)₂
2 : 1
0.0025 : 1/2×0.0025 = 0.00125
Mass of Ca(OH)₂:
Mass = number of moles × molar mass
Mass = 0.00125 mol × 74.1 g/mol
Mass = 0.09 g
Answer:
friction
Explanation:
The rubbing of certain materials against one another can transfer negative charges, or electrons.
