The mass of the piece of lead is calculated using the below formula
Q(heat)= mC delta T
Q = 78.0 j
M=mass =?
C=specific heat capacity ( 0.130 j/g/c
delat T=change in temperature = 9.0 c
by making M the subject of the formula
M = Q/ c delta T
M= 78.0 j/ 0.130 j/g/c x 9.0 c = 66.7 g of lead
Millimeters are an SI unit of length that =0.0001 m
Answer:
The mass of 10 cm³of a 0.4 g/dm³ solution of sodium carbonate is 0.004 grams
Explanation:
The question is with regards to density calculations
The density of the given sodium carbonate solution, ρ = 0.4 g/dm³
The volume of the given solution of sodium carbonate, V = 10 cm³ = 0.01 dm³


Therefore, we have;

The mass, "m", of the sodium carbonate in = ρ×V = 0.4 g/dm³ × 0.01 dm³ = 0.004 g
The mass of 10 cm³ (10 cm³ = 0.01 dm³) of a 0.4 g/dm³ solution of sodium carbonate, m = 0.004 g.
Hi there,
the answer to the blank is: boiling point
When a liquid is heated, the temperature stops rising at the liquid's boiling point.
Hope this is correct :)
Have a great day
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