Alright sorry you're getting the answer hours later, but i can help with this.
so you're looking for specific heat, the equation for it is <span>macaΔTa = - mbcbΔTb with object a and object b. that's mass of a times specific heat of a times final minus initial temperature of a equals -(mass of b times specific heat of b times final minus initial temperature of b)
</span>so putting in your values is, 755g * ca * (75 celsius - 84.5 celsius) = -(50g * cb * (75 celsius - 5 celsius))
well we know the specific heat of water is always 4180J/kg celsius, so put that in for cb
with a bit of simplification to the equation by doing everything on each side first you have, -7172.5 * ca = -14630000
divide both sides by -7172.5 so you can single out ca and you get, ca= 2039.74
add units for specific heat which are J/kg celsius and the specific heat of the material is 2039.74 J/kg celsius
Answer:
uh i think its D All of the above
Explanation:
sorry if its wrong
Since a percentage is out of 100, do the % / 100
Divide the percent by 100
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.
