Use the Ideal Gas Law to find the moles of gas first.
Be sure to convert T from Celsius to Kelvin by adding 273.
Also I prefer to deal with pressure in atm rather than mmHg, so divide the pressure by 760 to get it in atm.
PV = nRT —> n = PV/RT
P = 547 mmHg = 547/760 atm = 0.720 atm
V = 1.90 L
T = 33°C = 33 + 273 K = 306 K
R = 0.08206 L atm / mol K
n = (0.720 atm)(1.90 L) / (0.08206 L atm / mol K)(306 K) = 0.0545 mol of gas
Now divide grams by mol to get the molecular weight.
3.42 g / 0.0545 mol = 62.8 g/mol
Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
Answer:
c
Explanation:
the correct answer would be answer c
You need to know the energy frequency relationship for photons, which is thanks to Max Planck:
Photon Energy = Planck constant x Frequency
Rarranged:
Photon Energy / Planck Constant = Frequency
Planck Constant = 6.63x10^-34
2.93x10^-25 / 6.63x10^-34 = Frequency
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