Answer:
d. Electrons have a negative charge and are lighter than protons
Explanation:
The best statement that describes electrons is that electrons are negatively charged and are lighter than protons.
There are three known subatomic particles which are:
- Protons which are the positively charged particles within an atom.
- Electrons are the negatively charged particles
- Neutrons do not carry any charges.
The mass of the protons and neutrons are similar.
Electrons are lighter than both protons and neutrons.
Answer:
D if im pretty sure
Explanation:
In 1909 he began experiments that were to change the face of physics. He discovered the atomic nucleus and developed a model of the atom that was similar to the solar system.
20.06 g of Hg and 1.6 g of O₂
<u>Explanation:</u>
To Find:
Number of Mercury and oxygen that can be obtained from 21.7 g of HgO
First we have to write the balanced equation for the decomposition reaction of Mercury(II) oxide as,
2 HgO (s) → 2Hg(l) + O₂ (g)
21.7 g of HgO =
= 0.1 mol of HgO.
As per the above equation, we can find the mole ratio between HgO and Hg is 1: 1 and that of HgO and oxygen is 2:1 .
So amount of Hg produced = 0.1 mol × 200.59 g / mol ( molar mass of Hg)
= 20.06 g of Hg
Amount of oxygen produced = 0.05 mol × 32 g/ mol = 1.6 g of O₂
Thus it is clear that 20.06 g of Hg and 1.6 g of O₂ is obtained from 21.7 g of HgO
I think it’s radiation, that’s what i got from looking it up
When CH₄ is burnt in excess O₂ following products are formed,
CH₄ + 2 O₂ → CO₂ + 2 H₂O
According to equation 1 mole of CH₄ (16 g) reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O. Hence the products are,
1 mole of CO₂ and 2 moles of H₂O
Converting 1 mole CO₂ to grams;
As,
Mass = Moles × M.mass
Mass = 1 mol × 44 g.mol⁻¹
Mass = 40 g of CO₂
Converting 2 moles of H₂O to grams,
Mass = 2 mol × 18 g.mol⁻¹
Mass = 36 g of H₂O
Total grams of products;
Mass of CO₂ = 44 g
+ Mass of H₂O = 36 g
-------------
Total = 80 g of Product
Result:
80 grams of product is formed when 16 grams of CH₄ is burnt in excess of Oxygen.