Answer:
Explanation:
Oxidation state of Cr in CrPO₄
As a general rule, the algebraic sum of all the oxidation numbers of all atoms in a neutral compound is zero.
The compound above is in its neutral state and we sum all the oxidation numbers and equate to zero:
The oxidation number of P is -2
O is -2
Let the oxidation number of Cr be x:
x + (-2) + 4(-2) = 0
x -2-8 = 0
x -10 = 0
x = +10
For Cr in Cr₃(PO₄)₂
Using the same rule:
2(x) + 2[-2 + 4(-2)] = 0
2x + 2(-2-8) = 0
2x -20 = 0
x = +10
Answer:
349.22°C
Explanation:
Let the final temperature of the two pieces of metal be x.
Now, the warmer metal which is C u reduces from 475°C to x. Thus Δt for C u is; Δt1 = 475 - x.
The cooler metal Cr increases in temperature from 265°C to x. Thus, it's change in temperature is Δt for Cr is; Δt2 = x - 265.
Now from conservation of energy, the amount of energy leaving the C u metal is equal to the amount of energy entering the Cr metal.
Thus;
q_lost = q_gain
Where;
q_lost = m1•c1•Δt1
q_gained = m2•c2•Δt2
Now, c1 & c2 are the specific heat capacity of C u and Cr respectively.
From online tables, c1 = 0.385 J/g°C and c2 = 0.46 J/g°C
We are given;
m1 = 12g and m2 = 15g
Thus;
12 × 0.385 × (475 - x) = 15 × 0.46 × (x - 265)
2194.5 - 4.62x = 6.9x - 1828.5
6.9x + 4.62x = 2194.5 + 1828.5
11.52x = 4023
x = 4023/11.52
x = 349.22°C
Answer:
197mL of 0,506M HCl
Explanation:
The reaction of HCl + BaCO₃ is:
BaCO₃(s) + 2HCl → BaCl₂(aq) + CO₂ + H₂O.
The moles of BaCO₃ in 9,85 g are:
9,85 g of BaCO₃ × 
= <em>0,0499 moles of BaCO₃</em>
As 1 mol of BaCO₃ reacts with two moles of HCl, for a complete reaction of BaCO₃ to dissolve this compound in water you need:
0,0499 moles of BaCO₃ × 
=<em> 0,0998 moles of HCl</em>
If you have a 0,506M HCl, you need to add:
0,0998 moles of HCl× 
= 0,197 L ≡ 197mL
I hope it helps!
