The correct answer is the second option. Particulate matter is the air pollutant known to affect people with asthma. It also can cause other illnesses like cough, chest discomfort and a burning sensation in the lungs.
Answer: The standard state refers to 1 atm and
.
Explanation:
It is known that a chemical/substance can either be present in a solid, liquid or gaseous state.
So, when the phase of a substance like solid, liquid or gas is present at 1 atmosphere pressure and at a temperature of
then it known as standard state of substance.
Thus, we can conclude that standard state refers to 1 atm and
.
A because A is the only answer
Answer:
The answer to your question is below
Explanation:
a)
Number of atoms = ?
moles of Fe = 4.75
-Use proportions to solve this problem
1 mol of Fe --------------------- 6.023 x 10²³ atoms
4.75 moles --------------------- x
x = (4.75 x 6.023 x 10²³) / 1
x = 2.86 x 10²⁴ / 1
Number of atoms = 2.86 x 10²⁴
b)
Number of moles = ?
moles of 1.058 moles of H₂O
I think this question is incorrect, maybe you wish to know the number of atoms or grams of H₂O.
c)
Number of atoms = ?
moles of Fe = 0.759
1 mol of Fe ------------------ 6.023 x 10²³ atoms
0.759 moles --------------- x
x = (0.759 x 6.023 x 10²³) / 1
x = 4.57 x 10²³ / 1
Number of atoms of Fe = 4.57 x 10²³ atoms
d)
Number of molecules = ?
moles of H₂O = 3.5 moles
1 mol of H₂O ------------------ 6.023 x 10²³ molecules
3.5 moles ------------------ x
x = (3.5 x 6.023 x 10²³) / 1
x = 2.11 x 10²⁴ molecules
Number of molecules = 2.11 x 10²⁴
Answer:
3.65 x 10¹⁰ electrons
Explanation:
we'll apply the following equation for electric field of a point charge on a spherical conductor

where E is the electric field
k is a constant of the value 8.99 x 10⁹ Nm²/C²
r is the radius of the spherical conductor
q is the total charge in the sphere
Given diameter d =41.0cm, radius r = 20.5cm = 0.205m (convert cm to m)
Electrical field E = 1250 N/C
we are asked to determine how many excess electrons must be added to the surface of the sphere to produce this electric field

q = <u>E x r²</u>
k
q = <u>1250 N/C x 0.205m</u>²
8.99 x 10⁹ Nm²/C²
q = 5.84 x 10⁻⁹ C
this is the total charge in the sphere
To determine the number of electrons, we can divide the charge q by the charge on an electron e (1.6 x 10⁻¹⁹C)

n = <u>5.84 x 10⁻⁹ C </u>
1.6 x 10⁻¹⁹C
n = 3.65 x 10¹⁰ electrons
Therefore, to apply an electric field of magnitude 1250 N/C, the isolated spherical conductor must contain 3.65 x 10¹⁰ electrons