Answer:
Step-by-step explanation:
Remark
The formula for this is
Heat = m * c * Δt
Givens
Heat = unknown
m = 200 grams
c = 4.184 Joules / grams oC
Δt = the change in temperature = 35.7 - 22.3 = 13.4
Solution
Heat = 200 * 4.184 * 13.4
Heat = 11213 Joules
Heat = 11.2 Kj
It's hard to get the correct number of sig digs because the 200 is not qualified in any way.
Part 2.
This can really puzzle you until you know that heat given up = the heat taken on. The heat taken on was 11.2 Kj. The Butane lighter must have given that heat on.
Heat of Combustion = Heat Given Up / Number grams burned.
Heat of Combustion = 11.2 Kj / 0.23 grams
Heat of Combustion = 48.8 Kj / gram
Answer:
Equations 1 & 3 are correct.
Explanation:
Equation 2's actual answer is -8, and Equation 4's actual answer is positive 7.
Answer:
2.8a²+0.9a - 1.2
Step-by-step explanation:
Given the expression 0.3(3a-4)-0.05(8a)(-7a)
Expand using the distributive law
0.3(3a-4)-0.05(8a)(-7a)
0.3(3a)-0.3(4)-0.05(8a)(-7a)
0.9a-1.2 - 0.05(8a)(-7a)
0.9a - 1.2 + 2.8a²
Rearrange
2.8a²+0.9a - 1.2
Hence the required expression is 2.8a²+0.9a - 1.2
Solve by Elimination:
6y+5x=8
2.5x+3y=4
Multiply the second equations by 2:
5x+6y=8
we know see that both equations are the same line, this means that there is an infinite amount of solutions to the equation
(4 sec) * (-10 ft/sec) = -40 ft.
