Answer:
balanced in ACID not BASE
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
Answer
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
Explanation:
Cr2O7^2-(aq) + Hg(l) ----> Hg^2+(aqH) + Cr^3+(aq)
add H^1+ (acid) to capture the O and make 7 water molecules
Cr2O7^2-(aq) + Hg(l) + H^1+ ----> Hg^2+(aqH) + Cr^3+(aq) + 7H2O
Cr goes from +6 to +3 by gaining 3 e
Hg goes from 0 to +2 by losing 2 e
we need 3 Hg for every 2 Cr
so
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
2 Cr on the right and left
Net 12 positive charges on the right and the left
3 Hg on the right and left
14 H on the right and left
the equation is balanced
we cannot balance the equation in a basic solution with OH^1-
we have plenty of O in the dichromate ion. we need to convert it to water which take free H^1+ from the acid
Answer:
c) 3,3-dimethylpent-1-yne
A.Transform
B.Divergent
C. Convergent
40 g NaOH. You must use 40 g NaOH to prepare 10.0 L of a solution that has a pH of 13.
<em>Step 1</em>. Calculate the pOH of the solution
pOH = 14.00 – pH = 14.00 -13 = 1
<em>Step 2</em>. Calculate the concentration of NaOH
[NaOH] = [OH^(-)] = 10^(-pOH) mol/L = 10^(-1) mol/L = 0.1 mol/L
<em>Step 3</em>. Calculate the moles of NaOH
Moles of NaOH = 10.0 L solution × (0.1 mol NaOH/1 L solution) = 1 mol NaOH
<em>Step 4</em>. Calculate the mass of NaOH
Mass of NaOH = 1 mol NaOH × (40.00 g NaOH/1 mol NaOH) = 40 g NaOH
Answer:
The correct answer is -
1. a) The bubbles will shrink, some may vanish.
2. a) Can A will make a louder and stronger fizz than can B.
Explanation:
In the first question, it is given that the bottle is not opened and therefore, squeezing the bottle filled with a carbonated drink will increase the pressure on the carbonated liquid which forces the bubbles to dissolve or displace or vanish as it moves to empty space.
Thus, the correct answer would be - The bubbles will shrink, some may vanish
In the second question, there are two different conditions for two different unopened cans of carbonated water that are different temperatures one at the garage with higher temperature and one in the fridge at low temperature. As it is known that higher the temperature less will be solubility of gas in liquid so gas in can A will be less soluble which means it has more gas and it will make louder and stronger fizz than B which was stored at low temperature.
thus, the correct answer would be - Can A will make a louder and stronger fizz than can B.
