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.
Answer:
0.133 mL
Explanation:
Given data
- Initial concentration (C₁): 15.0 M
- Initial volume (V₁): to be determined
- Final concentration (C₂): 0.001 M
- Final volume (V₂): 2.00 L
We can find the volume of the concentrated solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.001 M × 2.00 L / 15.0 M
V₁ = 1.33 × 10⁻⁴ L = 0.133 mL
Answer:
Main-sequence stars, including the sun, form from clouds of dust and gas drawn together by gravity. ... The core that is left behind will be a white dwarf, a husk of a star in which no hydrogen fusion occurs. Smaller stars, such as red dwarfs, don't make it to the red giant state.
Explanation:
good luck i hope this helped!
Firstly the limiting reactant should be identified. Limiting reactant is the reactant that is in limited supply, the amount of product formed depends on the moles present of the limiting reactant.
the stoichiometry of x to y = 1:2
1 mole of x reacts with 2 moles of y
if x is the limiting reactant, there are 3 moles of x, then 6 moles of y should react, however there are only 4 moles of y. Therefore y is the limiting reactant and x is in excess.
4 moles of y reacts with 2 moles of x
since there are 3 moles of x initially and only 2 moles are used up, excess amount of x is 1 mol thats in excess.
