Answer:
d) carbon dioxide and water
We can't even see half of the question
First, we need to get the number of moles:
from the reaction equation when Y4+ takes 4 electrons and became Y, X loses 4 electrons and became X4+
∴ the number of moles n = 4
we are going to use this formula:
㏑K = n *F *E/RT
when K is the equilibrium constant = 4.98 x 10^-5
and F is Faraday's constant = 96500
and the constant R = 8.314
and T is the temperature in Kelvin = 298 K
and n is number of moles of electrons = 4
so, by substitution:
㏑4.98 x 10^-5 = 4*96500*E / 8.314*298
∴E = -0.064 V
Answer:
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Explanation:
Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.
H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
We will balance it using the trial and error method.
First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
Finally, we will get the balanced equation by multiplying H₂O by 6.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Answer is: mass of water is 56.28 grams.
Chemical reaction: 2H₂O → 2H₂ + O₂.
m(O₂) = 50.00 g.
n(O₂) = m(O₂) ÷ M(O₂).
n(O₂) = 50 g ÷ 32 g/mol.
n(O₂) = 1.5625 mol.
From chemical reaction: n(O₂) : n(H₂O) = 1 : 2.
n(H₂O) = 2 · 1.5625 mol.
n(H₂O) = 3.125 mol.
m(H₂O) = n(H₂O) · M(H₂O).
m(H₂O) = 3.125 mol · 18.01 g/mol.
m(H₂O) = 56.28 g.
