Endothermic reaction absorb energy leaving the product with more energy than the reactants.
Answer:
He must set the ball in motion by applying force in the desired direction.
Explanation:
That is how the soccer player will change the ball's position.
Have a good day :)
Answer:
you didnt give the question
Explanation:
<u>Answer:</u> The empirical and molecular formula of the compound is
and
respectively
<u>Explanation:</u>
We are given:
Mass of C = 3.758 g
Mass of H = 0.316 g
Mass of O = 1.251 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen = 
Moles of Oxygen = 
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.078 moles.
For Carbon = 
For Hydrogen = 
For Oxygen = 
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of C : H : O = 4 : 4 : 1
The empirical formula for the given compound is 
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is:

We are given:
Mass of molecular formula = 130 g/mol
Mass of empirical formula = 68 g/mol
Putting values in above equation, we get:

Multiplying this valency by the subscript of every element of empirical formula, we get:

Hence, the empirical and molecular formula of the compound is
and
respectively
There are 4 quantum numbers that can be used to describe the space of highest probability an electron resides in.
First quantum number is the principal quantum number- n , states the energy level.
Second quantum number states the angular momentum quantum number - l,
states the subshell and the shape of the orbital
values of l for n energy shells are from 0 to n-1
third is magnetic quantum number - m, which tells the specific orbital.
fourth is spin quantum number - s - gives the spin of the electron in the orbital
here we are asked to find l for 3p1
n = 3
and values of l are 0,1 and 2
for p orbitals , l = 1
therefore second orbital for 3p¹ is 1.