Soil temperature and <span>water content</span>
Answer:
v = 1130 cm³
Explanation:
Given data:
Volume of sample = ?
Mass of Al sample = 3.057 Kg (3.057 Kg× 1000g/1 Kg = 3057g)
Density of Al sample = 2.70 g/cm³
Solution:
Formula:
d = m/v
d = density
m = mass
v= volume
by putting values
2.70 g/cm³ = 3057g /v
v = 3057g /2.70 g/cm³
v = 1130 cm³
Answer:
2.7724 g
Explanation:
Mass of pre- 1892 pennies = 3.1 g
Abundance = 45.4 %
Mass of post 1892 pennies = 2.5 g
Abundance = 100 - 45.4 = 54.6 %
The average mass is given as = ( 3.1 g * 45.4 / 100) + (2.5g * 54.6 / 100)
Average Mass = 3.1 * 0.454 + 2.5 * 0.546
Average Mass = 1.4074 + 1.365 = 2.7724 g
Q = mcΔT = (4.00 g)(0.129 J/g•°C)(40.85 °C - 0.85 °C)
Q = 20.6 J of energy was involved (more specifically, 20.6 J of heat energy was absorbed from the surroundings by the sample of solid gold).
