Answer:
(x + 2)(x + 2)
Explanation:
You need 2 numbers that times to give 4 and add to give 4. So 2 and 2.
The one with 20 coils as it can generate more electricity.
<u>Answer:</u> The percentage abundance of
and
isotopes are 75.77% and 24.23% respectively.
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
Let the fractional abundance of
isotope be 'x'. So, fractional abundance of
isotope will be '1 - x'
- <u>For
isotope:</u>
Mass of
isotope = 34.9689 amu
Fractional abundance of
isotope = x
- <u>For
isotope:</u>
Mass of
isotope = 36.9659 amu
Fractional abundance of
isotope = 1 - x
- Average atomic mass of chlorine = 35.4527 amu
Putting values in equation 1, we get:
![35.4527=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7577](https://tex.z-dn.net/?f=35.4527%3D%5B%2834.9689%5Ctimes%20x%29%2B%2836.9659%5Ctimes%20%281-x%29%29%5D%5C%5C%5C%5Cx%3D0.7577)
Percentage abundance of
isotope = 
Percentage abundance of
isotope = 
Hence, the percentage abundance of
and
isotopes are 75.77% and 24.23% respectively.
Answer:
Explanation:
d =
m
V
m = d×V
V =
m
d
DENSITY
Density is defined as mass per unit volume.
d =
m
V
Example:
A brick of salt measuring 10.0 cm x 10.0 cm x 2.00 cm has a mass of 433 g. What is its density?
Step 1: Calculate the volume
V = lwh = 10.0 cm × 10.0 cm × 2.00 cm = 200 cm³
Step 2: Calculate the density
d =
m
V
=
433
g
200
c
m
³
= 2.16 g/cm³
MASS
d =
m
V
We can rearrange this to get the expression for the mass.
m = d×V
Example:
If 500 mL of a liquid has a density of 1.11 g/mL, what is its mass?
m = d×V = 500 mL ×
1.11
g
1
m
L
= 555 g
VOLUME
d =
m
V
We can rearrange this to get the expression for the volume.
V =
m
d
Example:
What is the volume of a bar of gold that has a mass of 14.83 kg. The density of gold is 19.32 g/cm³.
Step 1: Convert kilograms to grams.
14.83 kg ×
1000
g
1
k
g
= 14 830 g
Step 2: Calculate the volume.
V =
m
d
= 14 830 g ×
1
c
m
³
19.32
g
= 767.6 cm³
There is a shortcut trick while doing such fill in the blanks of nuclear reactions of hydrogen and helium
Let a,b,care elements of set N

Now
for our question
Hence b=4-3+1=1+1=2
So
The missing place should b e deuterium of heavy water
In nuclear reactions energy is released so it's mentioned on product side not reactant side