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
The average human has between sixty trillion and ninety trillion cells, but there are also organisms that have only a single cell. So basically it depends on the type of organism, but the minimum amount of cells is one.
Answer:
An insulated beaker with negligible mass contains liquid water with a mass of 0.205kg and a temperature of 79.9 °C How much ice at a temperature of −17.5 °C must be dropped into the water so that the final temperature of the system will be 31.0 °C? Take the specific heat for liquid water to be 4190J/Kg.K, the specific heat for ice to be 2100J/Kg.K, and the heat of fusion for water to be 334000J/kg.
The answer to the above question is
Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C
Explanation:
To solve this we proceed by finding the heat reaquired to raise the temperature of the water to 31.0 C from 79.9 C then we use tht to calculate for the mass of ice as follows
ΔH = m×c×ΔT
= 0.205×4190×(79.9 -31.0) = 42002.655 J
Therefore fore the ice, we have
Total heat = mi×L + mi×ci×ΔTi = mi×334000 + mi × 2100 × (0 -−17.5) = 42002.655 J
370750×mi = 42002.655 J
or mi = 0.1133 kg
Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C
The answer to this is true.
