The reaction between N₂ and F₂ gives Nitrogen trifluoride as the product. The balanced equation is;
N₂ + 3F₂ → 2NF₃
The stoichiometric ratio between N₂ and NF₃ is 1 : 2
Hence,
moles of N₂ / moles of F₂ = 1 / 2
moles of N₂ / 25 mol = 0.5
moles of N₂ = 0.5 x 25 mol = 12.5 mol
Hence N₂ moles needed = 12.5 mol
At STP (273 K and 1 atm) 1 mol of gas = 22.4 L
Hence needed N₂ volume = 22.4 L mol⁻¹ x 12.5 mol
= 280 L
Answer:
50000ppm and 0.855M.
Explanation:
ppm is an unit of chemistry defined as the ratio between mg of solute (NaCl) and Liters of solution. Molarity, M, is the ratio between moles of NaCl and liters
A 5% (w/v) NaCl contains 5g of NaCl in 100mL of solution.
To solve the ppm of this solution we need to find the mg of NaCl and the L of solution:
<em>mg NaCl:</em>
5g * (1000mg / 1g) = 5000mg
<em>L Solution:</em>
100mL * (1L / 1000mL) = 0.100L
ppm:
5000mg / 0.100L = 50000ppm
To find molarity we need to obtain the moles of NaCl in 5g using its molar mass:
5g * (1mol / 58.5g) = 0.0855moles NaCl
Molarity:
0.0855mol NaCl / 0.100L = 0.855M
The tool to measure the liquid is a measuring cylinder.
Sodium Sulfate
= Na2(SO4) meaning there are two ions of Na+ in one mole of Sodium Sulfate the M
stands for Molarity, defined as Molarity = (moles of solute)/(Liters of
solution), So if the Na2SO4 solution is 3.65M that means one Liter of has 3.65
moles of Na2SO4, the stoichiometry of Na2SO4 shows that there would be two Na+
ions in solution for every one Na2SO4.
Therefore if
3.65 moles of Na2SO4 was to dissolve, it would produce 7.3 moles of Na+, and
since this is still a theoretical solution, we can assume 1 L of solution.
Finally we find
[Na+] = 2*3.65 = 7.3M
Use the same
logic for parts b and c
