Number of coulombs of positive charge in 250cm^3 water is 1.3×10^7 C
The volume of 250 cm^3 corresponds to a mass of 250 g since the density of water is 1.0 g/cm^3
This mass corresponds to 250/18 = 14 moles since the molar mass of water is 18. There are ten proton (each with charge q = +e) in each molecule of 
So,
Q = 14NA q =14(6.02×10^23)(10)(1.60×10^−19C) = 1.3×10^7 C.
Mass is the quantity of matter in a physical body. It is also a measure of the body's inertia, the resistance to acceleration when a net force is applied. An object's mass also determines the strength of its gravitational attraction to other bodies.
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Answer:
The charge of an atom is the number of protons minus the number of electrons.
The Answer is B.
*NEGATIVE CHARGE, FOUND OUTSIDE THE NUCLEUS*
Explanation:
To answer this question, we'll need to use the Ideal Gas Law:
p
V
=
n
R
T
,
where
p
is pressure,
V
is volume,
n
is the number of moles
R
is the gas constant, and
T
is temperature in Kelvin.
The question already gives us the values for
p
and
T
, because helium is at STP. This means that temperature is
273.15 K
and pressure is
1 atm
.
We also already know the gas constant. In our case, we'll use the value of
0.08206 L atm/K mol
since these units fit the units of our given values the best.
We can find the value for
n
by dividing the mass of helium gas by its molar mass:
n
=
number of moles
=
mass of sample
molar mass
=
6.00 g
4.00 g/mol
=
1.50 mol
Now, we can just plug all of these values in and solve for
V
:
p
V
=
n
R
T
V
=
n
R
T
p
=
1.50 mol
×
0.08206 L atm/K mol
×
273.15 K
1 atm
= 33.6 L
this is not the answer but it will help you
do by the formula it is on the answer
Answer:
See explanation
Explanation:
For a reaction that proceeds by E1 mechanism, the rate determining step involves the formation of the carbocation.
The rate of formation of this carbocation depends only on the concentration of the t-butyl bromide since it is the only specie that enters into the rate equation.
Hence, when the concentration of t-butyl bromide is tripled, the rate of reaction is tripled.
Methanol does not enter into the rate equation hence doubling its concentration does not affect the rate of reaction.
