The answer is b) 6. An element in group 6A has 6 valence electrons. 2 valence electrons are in s orbital and 4 valence electrons are in p orbital.
Answer:
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Answer:
The answer to your question is a) limiting reactant SO₂
b)mass of SO₃ produced 31.25 g
Explanation:
Data
mass of SO₂ = 25 g
mass of O₂ = 40 g
Balanced Reaction
2SO₂ + O₂ ⇒ 2SO₃
Process
1.- Calculate the molecular mass of the reactants
SO₂ = 32 + (16 x 2) = 32 + 32 = 64
O₂ = 16 x 2 =32
2.- Use proportions to determine the limiting reactant
Theoretical proportion 2(SO₂) / O₂ = 2(64) / 32 = 4
Experimental proportion SO₂ / O₂ = 25 / 40 = 0.625
From these results, we determined that the limiting reactant is SO₂, because the proportion in the experiment is lower.
3.- Amount of SO₃ produced
Mass of SO₃ = 32 + (3 x 16) = 80 g
128 g of SO₂ --------------- 2(80) g of SO₃
25 g of SO₂ --------------- x
x = (25 x 2 x 80) / 128
x = 31.25 g of SO₃
Answer:
[KCl] = 1.2 M
Explanation:
We need to complete the reaction:
2KCl(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCl₂(s)↓
By stoichiomety we know that 1 mol of chloride needs 1 mol of nitrate to react:
Let's find out the moles of nitrate, we have:
Molarity = mol/volume(L)
We convert the volume → 30 mL . 1L/1000mL = 0.030L
Molarity . volume(L) = moles → 0.400 M . 0.030L = 0.012 moles
Therefore, we can make a rule of three.
1 mol of nitrate reacts with 2 moles of chloride
Then, 0.012 moles of nitrate must react with (0.012 . 2) / 1 = 0.024 moles of KCl
We convert the volume from mL to L → 20 mL . 1L /1000mL = 0.020L
Molarity = mol /volume(L) → 0.024 mol /0.020L = 1.2 M
