Answer:
For this angular momentum, no quantum number exist
Explanation:
From the question we are told that
The magnitude of the angular momentum is 
The generally formula for Orbital angular momentum is mathematically represented as
Where 
is the quantum number
now
We can look at the given angular momentum in this form as
comparing this equation to the generally equation for Orbital angular momentum
We see that there is no quantum number that would satisfy this equation
Answer:
Carbon dioxide is a product (cellular respiration)
Carbon dioxide is a reactant(photosynthesis)
Carried out in animals(cellular respiration)
Carried out in plants(both)
Chemical reaction(cellular respiration)
Oxygen is a product(photosynthesis)
Oxygen is a reactant(cellular respiration)
Produces usable energy source(photosynthesis)
Your welcome!!! Plz mark brainliest.
A match-head is flammable, but you have to get it up to its ignition temperature first.
A diamond is flammable, but its ignition temperature is astronomical and
totally out of sight.
Answer:
70 mL of 5% HCl and 30 mL of 15% HCl
Explanation:
We will designate x to be the fraction of the final solution that is composed of 5% HCl, and y to be the fraction of the final solution that is composed of 15% HCl. Since the percentage of the final solution is 8%, we can write the following expression:
5x + 15y = 8
Since x and y are fractions of a total, they must equal one:
x + y = 1
This is a system of two equations with two unknowns. We will proceed to solve for x. First, an expression for y is found:
y = 1 - x
This expression is substituted into the first equation and we solve for x.
5x + 15(1 - x) = 8
5x+ 15 - 15x = 8
-10x = -7
x = 7/10 = 0.7
We then calculate the value of y:
y = 1 - x = 1 - 0.7 = 0.3
Thus 0.7 of the 100 mL will be the 5% HCl solution, so the volume of 5% HCl we need is:
(100 mL)(0.7) = 70 mL
Similarly, the volume of 15% HCl we need is:
(100 mL)(0.3) = 30 mL
The distance of the earth to the sun in Mm = 1.5 x 10⁵
<h3>Further explanation</h3>
Given
The distance of the earth to the sun : 1.50 x 10⁸ km
Required
The distance in Mm
Solution
In converting units we must pay attention to the conversion factor.
the conversion factor :
1 kilometer(km) = 10⁻³ megameter(Mm)
So the distance conversion :
1.5 x 10⁸ x 10⁻³ = 1.5 x 10⁵ Mm
