It has the same composition throughout, unlike a heterogeneous which has visibly separate substances
Answer:
6.15.3 k
Explanation:
From the question we can see that
q = 0, Δu = w
Then,

putting values wet
=
T_f = 615.3 K
No He believed tiny particles were invisible and couldn't be changed....So No The person that believed in this was Dalton .
1) Temperature (heat) of the solution
2) Concentration (amount) of both solvent (usually water) and solute (substance being dissolved by solvent)
3) Movement (kinetic energy) of the solution, as in shaking/stirring
Answer:
7.98 × 10^3grams.
Explanation:
To find the mass of fluorine in the number of atoms provided, we first divide the number of atoms by Avagadros number (6.02 × 10^23atoms) to get the number of moles in the fluorine atom. That is;
number of moles (n) = number of atoms (nA) ÷ 6.02 × 10^23 atoms
n = 2.542 × 10^26 ÷ 6.02 × 10^23
n = 0.42 × 10^ (26-23)
n = 0.42 × 10^3
n = 4.2 × 10^2moles
Using mole = mass ÷ molar mass
Molar/atomic mass of fluorine (F) = 19g/mol
mass = molar mass × mole
Mass (g) = 19 × 4.2 × 10^2
Mass = 79.8 × 10^2
Mass = 7.98 × 10^3grams.