Answer:
20g
Explanation:
Given parameters:
Quantity of heat = 240calories
Initial temperature = 80°C
Final temperature = 68°C
Unknown:
Mass of water = ?
Solution;
To solve this problem, we use the expression:
Q = m c (t₂ - t₁)
Q is the quantity of heat
m is the mass
c is the specific heat = 1 cal/g°C
t₂ is the final temperature
t₁ is the initial temperature
Insert the parameters and solve;
Change in temperature is an absolute value
240 = m x 1 x (68 - 80)
m = 20g
Answer:
2 more neutrons
Explanation:
To obtain the answer to the question, let us calculate the number of neutrons in carbon–14 and standard carbon (i.e carbon–12). This can be obtained as follow:
For carbon–14:
Mass number = 14
Proton number = 6
Neutron number =?
Mass number = Proton + Neutron
14 = 6 + Neutron
Collect like terms
14 – 6 = Neutron
8 = Neutron
Neutron number = 8
For carbon–12:
Mass number = 12
Proton number = 6
Neutron number =?
Mass number = Proton + Neutron
12 = 6 + Neutron
Collect like terms
12 – 6 = Neutron
6 = Neutron
Neutron number = 6
SUMMARY:
Neutron number of carbon–14 = 8
Neutron number of carbon–12 = 6
Finally, we shall determine the difference in the neutron number. This can be obtained as follow:
Neutron number of carbon–14 = 8
Neutron number of carbon–12 = 6
Difference =?
Difference = (Neutron number of carbon–14) – (Neutron number of carbon–12)
Difference = 8 – 6
Difference = 2
Therefore, carbon–14 has 2 more neutrons than standard carbon (i.e carbon–12)
Answer:
640.59 g of NaCl.
Explanation:
Using Boiling point elevation,
ΔTb = kb * i * b
Where,
b = molality
kb = 0.52 °C kg/mole
i = van't koff factor
ΔTb = ΔTsol - ΔTsolv
= 2°C
2 = 0.52 * b
= 3.846 mol/kg.
Mass of water, solvent = 2.85 kg
Molality is the ratio of the number of moles of the solute to the mass of the solvent.
= 3.846 * 2.85
= 10.96 mol.
Mass of NaCl = molar mass * number of moles
= 58.44 * 10.96
= 640.59 g of NaCl.
Answer:
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Explanation:
Answer:
the static charge is not always distributed on the surface of the conductor, there are also charges in the volume but of lesser magnitude
Explanation:
In this hypothetical system the electric force is of type
F = 
in this case the force decays to zero much faster,
if we call Fo the force of Coulomb's law
F₀ = 
assuming the constant k is the same
the relationship between the two forces is
F / F₀ = 1 / r
F = F₀ / r
when analyzing this expression the force decays much faster to zero.
In an electric conductor, charges of the same sign may not feel any repulsive force from other charges that are at a medium distance, so there is a probability that some charges are distributed in the volume of the material, this does not happen with coulomb's law
Consequently, the static charge is not always distributed on the surface of the conductor, there are also charges in the volume but of lesser magnitude
