Answer:
3.01 × 10^24 atoms of vitamin D
Explanation:
The number of atoms, molecules or ions present in a substance is given by the Avogadro's number which is 6.02 × 10^23.
Hence;
1 molecule of vitamin D contains 6.02 ×10^23 atoms
5 molecules of vitamin D contains 5 × 6.02 ×10^23/1
= 3.01 × 10^24 atoms of vitamin D
694,563,239 rounded to the nearest thousand is 694,563.
It's because the first digit from the right is for ones, second for tens, third for hundreds and fourth for thousands and that's the one that we should take a closer look at. You can round it either to 3 or 4, depends on the digit of hundreds. In this case 3239 is clearly closer to 3000 than 4000, that's why we round it to 694,563, not 694,564.
Answer:
An atom with 1 valence electron and an atom with 7 valence electrons
Explanation:
Covalent bond:
It is formed by the sharing of electron pair between bonded atoms.
The atom with larger electronegativity attract the electron pair more towards it self and becomes partial negative while the other atom becomes partial positive.
For example:
In water the electronegativity of oxygen is 3.44 and hydrogen is 2.2. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and hydrogen becomes partial positive.
the number of valance electrons of oxygen are six and hydrogen is one that's why two hydrogen atoms are attached with one oxygen atom and complete the octet.
Stirring this is because the three elements are factors affecting dissolving of a solvent. Eg temprature affects in hotness or coldness, Particle size affects whether it is big or small while quantity of soluble affects by the amount
<u>Answer:</u> The value of
for the given reaction is 1.435
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:

Given mass of
= 9.2 g
Molar mass of
= 92 g/mol
Volume of solution = 0.50 L
Putting values in above equation, we get:

For the given chemical equation:

<u>Initial:</u> 0.20
<u>At eqllm:</u> 0.20-x 2x
We are given:
Equilibrium concentration of
= 0.057
Evaluating the value of 'x'

The expression of
for above equation follows:
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
![[NO_2]_{eq}=2x=(2\times 0.143)=0.286M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D2x%3D%282%5Ctimes%200.143%29%3D0.286M)
![[N_2O_4]_{eq}=0.057M](https://tex.z-dn.net/?f=%5BN_2O_4%5D_%7Beq%7D%3D0.057M)
Putting values in above expression, we get:

Hence, the value of
for the given reaction is 1.435