Answer : The concentration of luminol in the solution is, 1.20 M.
Explanation : Given,
Mass of luminol = 16.0 g
Volume of solution = 75.0 mL
Molar mass of luminol = 177.16 g/mole
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :

Now put all the given values in this formula, we get:

Therefore, the concentration of luminol in the solution is, 1.20 M.
O1Fl2
1. Assume an 100g sample, so the percentage will stay the same
2. Covert each element into their molar mass
29.6/16.00=1.8 mols of O
70.4/19.00=3.7 mols of Fl
3. Divide both by the smallest value of mol
1.8/1.8=1 O
3.7/1.8=2 Fl
4. Write the empirical formula:
O1Fl2
1) Write the balanced equation to state the molar ratios:
<span>3H2(g) + N2(g) → 2NH3(g)
=> molar ratios = 3 mol H2 : 1 mol N2 : 2 mol NH3
What volume of nitrogen is needed to produce 250.0 L of ammonia gas at STP?
First, convert the 250.0 L of NH3 to number of moles at STP .
Use the fact that 1 mole of gas at STP occupies 22.4 L
=> 250.0 L * 1mol/22.4 L = 11.16 L
Second, use the molar ratio to find the number of moles of N2 that produces 11.16 L of NH3
=> 11.16 L NH3 * [1 mol N2 / 2 mol NH3] = 5.58 mol N2
Third, convert 5.58 mol N2 into liters at STP
=> 5.58 mol N2 * [22.4 L/mol] = 124.99 liters
Answer: 124,99 liters
What volume of hydrogen is needed to produce 2.50 mol NH3 at STP?
First, find the number of moles of H2 that produce 2.50 mol by using the molar ratios:
2.50 mol NH3 * [3mol H2 / 2 mol NH3] = 3.75 mol H2
Second, convert the number of moles to liters of gas at STP:
3.75 mol * 22.4 L/mol = 84 liters of H2
Answer: 84 liters
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Answer:
From molar mass=total RAM of each individual element
78.8=(16+1)×3+M
78.8-51=M
27.8g/mol=M
Answer:
There are 1.05 x 10²⁴ molecules in 48.6 g N₂
Explanation:
1 mol of N₂ has a mass of (14 g * 2) 28 g.
Then, 48.6 g of N₂ will be equal to (48.6 g *(1 mol/ 28 g)) 1.74 mol.
Since there are 6.022 x 10²³ molecules in 1 mol N₂, there will be
(1.74 mol *( 6.022 x 10²³ / 1 mol)) 1.05 x 10²⁴ molecules in 1.74 mol N₂ (or 48. 6 g N₂).