Answer:
A. 2Cu(s) + S(s) → Cu₂S(s)
B. 2C₈H₁₈(g) + 25O₂(g) → 16CO₂(g) + 18H₂O(g)
C. 2SO₂(g) + O₂(g) → 2SO₃(g)
Explanation:
A. Let's think the reactants:
Cu and S
The product is: CuS
Equation: 2Cu(s) + S(s) → Cu₂S(s)
B. The reactants are:
O₂ and C₈H₁₈
The products are: water and CO₂
This is a combustion reaction: 2C₈H₁₈(g) + 25O₂(g) → 16CO₂(g) + 18H₂O(g)
C. Reactants: SO₂ and O₂
Products: SO₃
Equation: 2SO₂(g) + O₂(g) → 2SO₃(g)
The water cycle renews water supplies by moving water from the oceans, as freshwater, to land
Answer:
C4H4O
Explanation:
1.24 g x 1mol/(44)g = 0.03mol C;
Mass of C: 0.03mol x 12.g/1mol
= 0.36g of C
mol of H: 0.255g x 1mol /(18) x2= 0.028mol
Mass of H; = 0.028g × 1 of H
Mass of O: 0.519 g – (0.36g +0.028g)= 0.131g;
0.131g/16 =0.008 mol
Divide by the smallest number of moles;
0.028/0.008 mol = 3.5; 0.03/0.008 mol =3.75, 0.008 mol/0.008 mol =1
C4H4O
Answer:
The correct answer is - C) 3 hours.
Explanation:
Given:
The half-life = 1 hour
Solution:
The half-life of an element is the time required to decay half of its initial amount.
To complete 87.5%, there total numbers of half-live would be -
1st halflife = 50% of the initial amount
2nd half-life = 50 + 50% of (50) = 50 +25 = 75% of initial amount
3rd half-life = 50 + 50 % of (50) + 50% of (25) = 87.5% of initial amount
so, to complete the 87.5 % of decay it required 3 half lives, and each half-life take one hour thus, total time taken = 3 hours
Answer:
oxidation reaction.
Explanation:
Every reduction reaction must be accompanied by an oxidation reaction.
Oxidation leads electrons loss whereas reduction implies gain of electrons. So missing electrons should always be the equivalent of obtained electrons. Without something acquiring electrons there can't be any loss. Electrons can't simply disappear!
It implies, but, that oxidation and reduction must occur simultaneously at different locations and the electrons can commute across cables or in a liquid medium by ions.