D. Making the reactant particles larger
Answer:
T2 = 29.79°C
Explanation:
Equliibrium signifies that heat loss = heat gained
Heat gained by Ice;
H = ML
Mass, M = Number of moles * Molar mass = 1 * 18 = 18g
l = 6.01 k J m o l = 334 J/g
C = 4.186 J/g
H = 18(334)
H = 6012
Heat lost by water
H = MCΔT
H = 18 * 4.186 * (50 - T2)
H = 3767.4 - 75.348T2
Since H = H, we have;
6012 = 3767.4 - 75.348T2
- 75.348T2 = 3767 - 6012
T2 = 2245 / 75.348
T2 = 29.79°C
Answer:
Explanation:
We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
You don't tell us what the reaction is, but we can solve the problem so long as we balance the OH.
M_r: 58.32
Mg(OH)₂ + … ⟶ … + 2HOH
m/g: 58.3
(a) Moles of Mg(OH)₂
(b) Moles of H₂O
The molar ratio is 2 mol H₂O = 1 mol Mg(OH)₂.
The reaction will form 
of water.
Answer:
0.00050553
Explanation:
when the power of ten is negative, move the decimal to the left
hope this helped!
Reduction half reaction: Cu²⁺(aq) + 2e⁻ → Cu⁰(s).
Oxidation half reaction: NO₂⁻(aq) + H₂O(l) → NO₃⁻(aq) + 2H⁺(aq) + 2e⁻.
Balanced chemical reaction:
Cu²⁺(aq) + NO₂⁻(aq) + H₂O(l) → Cu(s) + NO₃⁻(aq) + 2H⁺(aq).
Copper is reduced from oxidation number +2 (Cu²⁺) to oxidation number 0 (Cu) and nitrogen is oxidized from oxidation number +3 (in NO₂⁻) to oxidation number +5 (in NO₃⁻).
