H2O2(I)
C6H6(O)
CO2(I)
C2H6(O)
HNO3(I)
Answer:
An aldose is defined as a monosaccharide whose carbon skeleton has an aldehyde group. Ketose is a monosaccharide whose carbon skeleton has a ketone group.
Explanation:
Answer:
The amount of salt in the tank at the end of an additional 10 minutes are 4.38lb.
Explanation:
<u>Situation 1:</u>
Tank with 100 gallons of fresh water
<u>Situation 2:</u>
Tank with 100 gallons of fresh water + water with 0.5lb of salt per gallon
After 10 minutes, as the rate in which the new water is poured is 2 gallons per minute, the result is 20 gallons added (2×10=20) . And taking in account that the water contains 0.5 lb of salt per gallon the amount of salt added is 20×0.5= 10lb of salt.
That amount of salt is now in all the water inside the tank which is 100 gallons+ 20 gallons= 120 gallons. <em>That means that in situation 2 we have 10lb of salt in 120 gallons of water.</em>
That mixture is allowed to leave the tank at a rate of 2 gallons per minute so we will have after 10 minutes: 120 gallons- (2×10) gallons= 100 gallons remaining in the tank. And the amount of salt if we remember that we had 10lb in 120 gallons, now in 100 gallons we will have: (100 gallons × 10lb of salt)/ 120 gallons= 8.33 lb of salt.
<u>Situation 3:</u>
Tank with 100 gallons of water with 8.33lb of salt.
After 10 minutes in which fresh water is poured in the tank at a rate of 9 gallons per minute, the result is: 9×10= 90 gallons added to the tank. So now we have 100+90=190 gallons of water in the tank. <em>That means in situation 3 we have 8.33 lb of salt in 190 gallons of water. </em>
That mixture is leaving the tank at a rate of 9 gallons per minute so we have after 10 minutes: (190- (9×10))= 100 gallons of mixture remaining in the tank.
And the amount of salt if we remember that we had 8.33lb in 190 gallons, now in 100 gallons we will have: (100 gallons × 8.33lb of salt)/ 190 gallons= 4.38 lb of salt.
The details of the galvanic cells are missing so i have attached it.
Answer:
The voltage will increase in cell Y and decrease in cell Z.
Explanation:
Looking at the reaction, in half cell 3, we see that reduction takes place in galvanic cell Y while oxidation takes place in galvanic cell Z.
In galvanic cell Y, aluminum gains electrons which is reduction while in galvanic cell Z, Copper loses electrons which is oxidation.
Now, electricity involves movement of electrons and where there is a loss of electrons, it means a decrease in electricity and when there is a gain in electrons, it means there is an increase in electricity.
Now, in electricity, Voltage is directly proportional to electric power.
Thus, the higher the voltage, the higher the electric power.
Thus, in galvanic cell Y that undergoes gain in electrons, there will be an increase in voltage while in galvanic cell Z that undergoes loss of electrons, there will be a decrease in voltage.