Answer:
Gold
Explanation:
We are given that
Mass of sample ,m=385 g
Volume ,V=20mL
We have to find the coin is gold or yellow brass.
We know that
Density=
Using the formula
Density of coin=19.3g/mL
Density of gold=19.3g/mL
Hence, the coin is gold.
Missing in your question Ka2 =6.3x10^-8
From this reaction:
H2SO3 + H2O ↔ H3O+ + HSO3-
by using the ICE table :
H2SO3 ↔ H3O + HSO3-
intial 0.6 0 0
change -X +X +X
Equ (0.6-X) X X
when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088
by using the ICE table 2:
HSO3- ↔ H3O + SO3-
initial 0.088 0.088 0
change -X +X +X
Equ (0.088-X) (0.088+X) X
Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] = 0.088 as the value of Ka2 is very small
6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
= 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
= -㏒ 0.088 = 1.06
Answer:
a push
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Explanation:
Answer:
126.0g of water were initially present
Explanation:
The electrolysis of water occurs as follows:
2H₂O(l) ⇄ 2H₂(g) + O₂(g)
<em>Where 2 moles of water produce 2 moles of hydrogen and 1 mole of oxygen.</em>
<em />
To find the mass of water we need to determine moles of oxygen and hydrogen, thus:
<em>Moles Hydrogen:</em>
14.0g H₂ ₓ (1mol / 2g H₂) = 7 moles H₂
<em>Moles Oxygen:</em>
112.0g O₂ ₓ (1mol / 32g) = 3.5 moles O₂
Based on the chemical equation, the moles of water initially present were 7 moles (That produce 7 moles H₂ and 3.5 moles O₂). The mass of 7 moles of H₂O is:
7 moles H₂O * (18g / mol) =
<h3>126.0g of water were initially present</h3>
