Answer:
A. False.
Every substance contains the same number of molecules i.e 6.02x10^23 molecules
B. False.
Mass conc. = number mole x molar Mass
Mass conc. of 1mole of N2 = 1 x 28 = 28g
Mass conc. of 1mol of Ar = 1 x 40 = 40g
The mass of 1mole of Ar is greater than the mass of 1mole of N2
C. False.
Molar Mass of N2 = 2x14 = 28g/mol
Molar Mass of Ar = 40g/mol
The molar mass of Ar is greater than that of N2.
Explanation:
Answer:
I'm pretty sure that's right.
Explanation:
Answer:
The answer to your question is 80.3%
Explanation:
Data
Percent by mass of F
molecules NF₃
Process
1.- Calculate the molar mass of nitrogen trifluoride
molar mass = (1 x 14) + (19 x 3)
= 14 + 57
= 71 g
2.- Use proportions and cross multiplications to find the percent by mass of F. The molar mass of NF₃ is equal to 100%.
71 g of NF₃ ------------------ 100%
57 g of F ------------------- x
x = (57 x 100)/71
x = 5700 / 71
x = 80.3%
3.- Conclusion
Fluorine is 80.3% by mass of the molecule NF₃
The burning of methane gas, given below, is a redox reaction. which part of the reaction illustrates oxidation, Ch4+O2---CO2+H2O<span>CH4---CO2</span>
Answer:
Explanation:
Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts with oxygen. What is the limiting reactant in this experiment?
Mg + O2 → MgO (unbalanced)
first, balance the equation
2Mg +O2-------> 2MgO
two magnesium atoms react with one diatomic oxygen molecule
there is a 1:1 ratio of magnesium to oxygen atoms
but we have 2 moles of magnesium atoms and 2X5 = 10 moles of oxygen atoms
the lesser magnesium LIMITS the amount of product we can make, so it is the LIMITING REAGENT.
