Answer:
#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.
Explanation:
Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆
1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.
=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP
2. Calculate moles of F₂ used
=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used
3. Calculate moles of XeF₆ produced from reaction ratios …
Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆
4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number (6.02 x 10²³ molecules/mole)
=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)
= 2.75 x 10²³ molecules XeF₆.
The last set which is n=4 l=3 m=3 is a valid set
Answer:
33.33% = 33%
Explanation:
MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g)
1 mole of MCO3 will produce → 1 mole of CO2
We need to get the number of mole of CO2:
and when we have 0.22 g of CO2, so number of mole = mass / molar mass
Moles = 0.22 g / 44 g/mol = 0.005 mole
Moles of Mg = moles of CO2 = 0.005 mole
Mass of Mg = moles * molar mass
= 0.005 * 84 /mol = 0.42 g
Percent of MgCO3 by mass of Mg = 0.42 g / 1.26 * 100
=33.33 %
The number of protons in the nucleus of the atom is equal to the atomic number (Z). The number of electrons in a neutral atom is equal to the number of protons. The mass number of the atom (M) is equal to the sum of the number of protons and neutrons in the nucleus.
It’s lower, hope this helps with your issue and helps solve your problem, no problem
