7. If 294 J of heat is transferred to a 10.0 g sample of silver at 25 ºC, what is the final temperature of the silver? Specific heat capacity of silver is 0.235 J/gºC.

T~final is 150.11 C

Step-by-step explanation:

To solve this we need this specific heat equation:

Q=mc(T~final - T~initial) that is the difference of final temperature minus initial temperature.

We are given:

Q = 294 J

m = 10 g

c = 0.235 J/gC

T~initial = 25 C

Plug the values in to the equation

294 J = 10 g * 0.235 J/gC * T(difference)

294 J = 2.35 J/C * T(difference)

Divide both sides by 2.35 J/C

294 J / 2.35 J/C = T(difference)

T(difference) = 125.11 C

If the initial temperature was 25 C then then

T~final = 25 C + 125.11 C or 150.11 C

the camry sales+accord sale=828,000. to make it so the two numbers are 18 thousand apart and still equal 828,000, you first need to find out what half of 828,000 is. 828,000/2=414,000. for the two numbers to equal 828,000, you have to add the same amount as you subtract from the middle. in this case, the number is 9,000 because it's half of 18,000. to equal 828,000, you have to add 9,000 to 414,000 for the camry and subtract 9,000 from 414,000 for the accord. 414,000+9,000=423,000.

414,000-9,000=405,000.

423,000+405,000=828,000.

423,000-405,000=18,

hope it

total of apples - take away four

6-4 = 2

answer is 2 apples