The solution would be like
this for this specific problem:
<span>Moles of carbon = 58.8 /
12 = 4.9 </span><span>
<span>Moles of hydrogen = 9.8 / 1 = 9.8 </span>
<span>Moles of oxugen = 31.4 / 16 m= 1.96 </span>
<span>Ratio 4.9 / 1.96 = 2.5 9.8 / 1.96 = 5.0 1.96 / 1.96 = 1 </span></span>
Simplest
formula = C5H10<span>
</span><span>I hope this helps and if
you have any further questions, please don’t hesitate to ask again.</span>
Answer:
Explanation:
Iron is Fe, nitrogen is N. Nitrogen is diatomic, which means it occurs as a molecular pair by itself. Iron III nitride has a chemical formula of FeN because nitrogen has a charge of 3-, and iron III tells us the iron has a charge of 3+ so you just need one of each to make the charges balance and the compound neutral.
Answer:
0.109 g.
Explanation:
Equation of the reaction:
Na3PO4 + 3HCl --> 3NaCl + H3PO4
Number of moles of HCl = molar concentration × volume
= 0.1 × 0.04
= 0.004 mol.
By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3
= 0.0013 mol
Mass of Na3PO4 = molar mass × number of moles
= 0.0013 × 164
= 0.219 g
Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance
= 0.219 × 50 g/100 g
= 0.109 g.
Explanation:
option A skeleton system is correct option
hope this helps you !
So the first thing we must do is write a balanced equation for the reaction and we know the equation is balnced when all the species on the RHS is equal to the species on the LHS
2NaOH + H₂SO₄ → Na₂SO₄<span>
+ 2H₂O</span>
So now it's time to identify what reactant you know the most for from the question (volume & conc. of H₂SO₄) and use that info to find the unknown (conc. of NaOH)
If 1000 ml of H₂SO₄ contain 0.750 mol [0.750 M is the amount of moles in
1 L (1000 ml)]
then let 15 ml of H₂SO₄ contain x mol [15 ml is the amount of the acid that took part in the reaction]
⇒
x = = 0.01125 molMole ratio of NaOH to H₂SO₄ can be obtained from the balanced equation
0
2NaOH +
1H₂SO₄ → Na₂SO₄ + 2H₂O
mole ratio of NaOH to H₂SO₄ is 2 : 1∴ if mole of of H₂SO₄ = 0.01125 mol then moles of NaOH = (0.01125 mol) × 2 = 0.0225 molIf 17.5 ml of NaOH contain 0.0225 mol [this was given in the question]
then let 1000 ml of NaOH contain x⇒ x = = 1.286 mol∴ concentration of NaOH is 1.286 mol/L