Given:
AgCl (s) ===> Ag+ (aq) + Cl- (aq)
- negative entropy
H2O(g) ===> H2O(l)
- positive entropy
2Na (s) + Cl2 (g) ===> 2NaCl (s)
- positive entropy
Br2(l) ===> Br2 (s)
-positive entropy
They are identified to have positive or negative values of entropy based on the phases of the reactants and products.
Answer:
7.] 597.11 Calories
8] 50208 joules
9] 2.05016 kilojoules
10] 0.0142256 kilojoules
Explanation:
just google how to do it. you can find some good examples.
hope this helps a bit!
brainliest?
also, im in 7th, im in high school algebra already. (im in advanced stuff)
ty!
Answer:
Alkali metals are highly reactive elements that appear to be silver and they are found in group 1 of the periodic table. It consists of lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr). As you go further down the group, the more reactive they are. Those elements all react to water and air, so they must be kept in oil to preserve their state.
Answer:
M = 0.441 M
Explanation:
In this case, we have two solutions that involves the Manganese II cation;
We have Mn(CH₃COOH)₂ and MnSO₄
In both cases, the moles of Mn are the same in reaction as we can see here:
Mn(CH₃COO)₂ <-------> Mn²⁺ + 2CH₃COO⁻
MnSO₄ <------> Mn²⁺ + SO₄²⁻
Therefore, all we have to do is calculate the moles of Mn in both solutions, do the sum and then, calculate the concentration with the new volume:
moles of MnAce = 0.489 * 0.0283 = 0.0138 moles
moles MnSulf = 0.339 * 0.0125 = 0.0042 moles
the total moles are:
moles of Mn²⁺ = 0.0138 + 0.0042 = 0.018 moles
Finally the concentration: 12.5 + 28.3 = 40.8 mL or 0.0408 L
M = 0.018 / 0.0408
M = 0.441 M
This would be the final concentration of the manganese after the mixing of the two solutions
