In this question, you are given the wine/alcohol concentration (12%) and the volume of the alcohol(0.75l). You are asked the alcohol content in milli liters, so don't forget to change the unit. Then the equation would be:
Alcohol content= alcohol concentration * bottle volume
Alcohol content = 12%* 0.75l * 1000ml/l= 90ml
Answer:
1.39
Explanation:
[Hg2Cl2]= 1M
[H^+] = ????
E°cell= 0.35V
E= 0.268 V
Therefore E for the reaction must -0.082 V
n= 2 moles of electrons
From Nernst Equation:
E= E°cell- 0.0592/n log [Red]/[Ox]
0.0268= 0.35- 0.0592/2 log 1/[Ox]^2
-0.082= -0.0296 log 1/[Ox]^2
log 1/[Ox]^2= 0.082/0.0296
log 1/[Ox]^2= 2.77
1/[Ox]^2=Antilog (2.77)
[Ox]^2=1.698×10^-3
[Ox] = 0.0412 M
But pH= -log [H^+]= -log(0.0412)= 1.385
Answer:
<em> ionic equation : </em>3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)
<em> net ionic equation: </em>3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)
Explanation:
The balanced equation is
3FeSO4(aq)+ 2Na3PO4(aq) → Fe3(PO4)2(s)+ 3Na2SO4(aq)
<em>Ionic equations: </em>Start with a balanced molecular equation. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions
. Indicate the correct formula and charge of each ion. Indicate the correct number of each ion
. Write (aq) after each ion
.Bring down all compounds with (s), (l), or (g) unchanged. The coefficents are given by the number of moles in the original equation
3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)
<em>Net ionic equations: </em>Write the balanced molecular equation. Write the balanced complete ionic equation. Cross out the spectator ions, it means the repeated ions that are present. Write the "leftovers" as the net ionic equation.
3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)
Answer:
THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K
Explanation:
Mass of alloy = 33 g
Initial temperature of alloy = 93°C
Mass of water = 50 g
Initail temp. of water = 22 °C
Heat capacity of calorimeter = 9.20 J/K
Final temp. = 31.10 °C
specific heat of alloy = unknown
specific heat capacity of water = 4.2 J/g K
Heat = mass * specific heat * change in temperature = m c ΔT
Heat = heat capcity * chage in temperature = Δ H * ΔT
In calorimetry;
Heat lost by the alloy = Heat gained by water + Heat of the calorimeter
mc ΔT = mcΔT + Heat capacity * ΔT
33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)
33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1
2042.7 C = 1911 + 83,72
C = 1911 + 83.72 / 2042.7
C = 1994.72 /2042.7
C =0.9765 J/g K
The specific heat of the alloy is 0.9765 J/ g K
A weak Bronsted-Lowry base is a weak proton acceptor, where the proton is in the form of H+, so the conjugate acid formed contains one more H atom and an extra positive charge.
Hope this helps!
