Answer:
2.77 mL of boiling water is the minimum amount which will dissolve 500 mg of phthalic acid.
Explanation:
We know from the problem that 18 g of phthalic acid are dissolved in 100 mL of water at 99 °C.
Now we devise the following reasoning:
If 18 g of phthalic acid are dissolved in 100 mL of water at 99 °C
Then 0.5 g of phthalic acid are dissolved in X mL of water at 99 °C
X = (0.5 × 100) / 18 = 2.77 mL of water
Answer:
The change in temperature that occurs when 8000 J of heat is used by a mass 75 g of water is 25.4 °C
Explanation:
H = mc ΔT
m = 75 g
c = 4. 200 J/ g °C
H = 8000 J
ΔT =?
Rearranging the formula, making ΔT the subject of formula;
ΔT = H / m c
ΔT = 8000 / 75 * 4.200
ΔT = 8000 / 315
ΔT = 25.4 °C
There are no states in the picture, but Na should have a "(s)" after it, and Cl2 should have a "(g)" after it. NaCl should have an "(s)". Chlorine is a diatomic element so it has a "2" subscript on it.
Hope this helped! :)
Answer:
-3.7771 × 10² kJ/mol
Explanation:
Let's consider the following equation.
3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)
We can calculate the standard Gibbs free energy (ΔG°) using the following expression.
ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)
where,
n: moles
ΔG°f(): standard Gibbs free energy of formation
p: products
r: reactants
ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))
ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)
ΔG° = -377.71 kJ = -3.7771 × 10² kJ
This is the standard Gibbs free energy per mole of reaction.
Answer:
A. Yes, the substance must be water.
Explanation:
The density of a substance is unique to it. Density is defined the as the amount of substance contained per volume.
One of the ways of identifying a substance is to determine its density. Every matter is known to have their own specific densities. This makes them different from other substances. The density of gold is unique to it and it differs from that of silver.
In fact, water has density of 1.00gcm⁻³. Experimental errors and some little factors must have altered our expected figure. This a case of precision and accuracy in the experiment.
