<span>Answer:
Graham's law of gaseous effusion states that the rate of effusion goes by the inverse root of the gas' molar mass.
râšM = constant
Therefore for two gases the ratio rates is given by:
r1 / r2 = âš(M2 / M1)
For Cl2 and F2:
r(Cl2) / r(F2) = âš{(37.9968)/(70.906)}
= 0.732 (to 3.s.f.)</span>
Answer:
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the required grams of HCl by firstly identifying the limiting reactant via the moles of each reactant as they are in a 1:1 mole ratio:
Thus, we infer the hydrogen is the limiting reactant and therefore we use its 1:2 mole ratio with HCl whose molar mass is 36.46 g/mol:
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Answer:
Part A = The mass of sulfur is 6.228 grams
Part B = The mass of 1 silver atom is 1.79 * 10^-22 grams
Explanation:
Part A
Step 1: Data given
A mixture of carbon and sulfur has a mass of 9.0 g
Mass of the product = 27.1 grams
X = mass carbon
Y = mass sulfur
x + y = 9.0 grams
x = 9.0 - y
x(molar mass CO2/atomic mass C) + y(molar mass SO2/atomic mass S) = 22.6
(9 - y)*(44.01/12.01) + y(64.07/32.07)
(9-y)(3.664) + y(1.998)
32.976 - 3.664y + 1.998y = 22.6
-1.666y = -10.376
y = 6.228 = mass sulfur
x = 9.0 - 6.228 = 2.772 grams = mass C
The mass of sulfur is 6.228 grams
Part B
Calculate the mass, in grams, of a single silver atom (mAg = 107.87 amu ).
Calculate moles of 1 silver atom
Moles = 1/ 6.022*10^23
Moles = 1.66*10^-24 moles
Mass = moles * molar mass
Mass = 1.66*10 ^-24 moles *107.87
Mass = 1.79 * 10^-22 grams
The mass of 1 silver atom is 1.79 * 10^-22 grams
Answer: After 4710 seconds, 1/8 of the compound will be left
Explanation:
Using the formulae
Nt/No = (1/2)^t/t1/2
Where
N= amount of the compound present at time t
No= amount of compound present at time t=0
t= time taken for N molecules of the compound to remain = 4710 seconds
t1/2 = half-life of compound = 1570 seconds
Plugging in the values, we have
Nt/No = (1/2)^(4710s/1570s)
Nt/No = (1/2)^3
Nt/No= 1/8
Therefore after 4710 seconds, 1/8 molecules of the compound will be left
Answer:
B. Metal
Explanation:
The catalyst used in the converter is mostly a precious metal such as platinum, palladium and rhodium. Platinum is used as a reduction catalyst and as an oxidation catalyst. Although platinum is a very active catalyst and widely used, it is very expensive and not suitable for all applications.
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