Answer:
0.082g
Explanation:
The following data were obtained from the question:
Heat (Q) = 0.092J
Change in temperature (ΔT) = 0.267°C
Specific heat capacity (C) of water = 4.184J/g°C
Mass (M) =..?
Thus, the mass of present can be obtained as follow:
Q = MCΔT
0.092 = M x 4.184 x 0.267
Divide both side by 4.184 x 0.267
M = 0.092 / (4.184 x 0.267)
M = 0.082g
Therefore, mass of water was present is 0.082.
Nonmetal your welcome
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7 different types of tide
Density = 0.601 g/mL
Volume = 7.25 mL
D = m / V
0.601 = m / 7.25
m = 0.601 x 7.25 => 4.35725 g
hope this helps!
