The correct option is B.
Isotopes refers to those chemical compounds which have the same number of protons and electrons but different number of neutrons, so they end up having different mass numbers. The diagram given above is that of beryllium, which has atomic number 4 and it has 2 electrons in its outermost shell. It has four protons [same as the number of electrons] and 5 neutrons. Beryllium 10, which is its isotope has four electrons, four protons and 6 neutrons. To get the number of neutron, remove the number of electrons from the number given in the option, that is, 10 - 4 = 6.
Answer:
a) 1.248 x 10⁷ kg
b) 1.248 x 10⁴ Mg
c) 1.248 x 10¹³ mg
d) 1.248 x 10⁴ ton
Explanation:
a) Since 1000 g = 1 kg we can convert grams to kg by multiplyig any given quantity in grams by the conversion factor ( 1 kg / 1000 g):
1.248 x 10¹⁰ g * (1 kg / 1000 g) = 1.248 x 10⁷ kg
b) Since 1 Mg = 1 x 10⁶ g, the conversion factor will be ( 1 Mg / 1 x 10⁶ g):
1.248 x 10¹⁰ g * ( 1 Mg / 1 x 10⁶ g) = 1.248 x 10⁴ Mg
c) Since 1 mg = 1 x 10⁻³ g, the conversion factor will be ( 1 mg / 1 x 10⁻³ g):
1.248 x 10¹⁰ g ( 1 mg / 1 x 10⁻³ g) = 1.248 x 10¹³ mg
d) Since 1 metric ton = 1000 kg and 1000 g = 1 kg, we can use these conversions factors: ( 1 kg / 1000 g) and (1 ton / 1000 kg):
1.248 x 10¹⁰ g * ( 1 kg / 1000 g) * ( 1 ton / 1000 kg) = 1.248 x 10⁴ ton
Answer:
Avogadro number of representatives particles is equal to one mole.
Explanation:
The number 6.022 × 10²³ is called Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
17 g of ammonia = 1 mole = 6.022 × 10²³ molecules of ammonia
12 g of carbon = 1 mole = 6.022 × 10²³ atoms of carbon
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
Answer: im thinking its gonna be d.C2H6 and also
the explanation is on the research i had did before i had answered this question so i really hope this help :)
Explanation:
Ar = van de waals forces or london forces
C
H
4
= van de waals forces or london forces
HCl=permanent dipole-dipole interactions
CO = permanent dipole-dipole interactions
HF = hydrogen bonding
N
a
N
O
3
= permanent dipole-dipole interactions
C
a
C
l
2
= van de waals forces or london forces
