Answer:

Explanation:
You don't give the reaction, but we can get by just by balancing atoms of Na.
We know we will need the partially balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.
M_r: 142.04
2NaOH + … ⟶ Na₂SO₄ + …
n/mol: 0.75
1. Use the molar ratio of Na₂SO₄ to NaOH to calculate the moles of NaF.
Moles of Na₂SO₄ = 0.75 mol NaOH × (1 mol Na₂SO₄/2 mol NaOH
= 0.375 mol Na₂SO₄
2. Use the molar mass of Na₂SO₄ to calculate the mass of Na₂SO₄.
Mass of Na₂SO₄ = 0.375 mol Na₂SO₄ × (142.04 g Na₂SO₄/1 mol Na₂SO₄) = 53 g Na₂SO₄
The reaction produces
of Na₂SO₄.
Answer:
1. No
2.a. Nothing will happen to figure 1 as both the sides have 30 N.
2.b. The force with 30 N will push 10 N because 10 N is less force than 30 N.
Answer:
A working model used to test a design is called a
stop
Answer:
Explanation:
Before Hybridization => C: 1s²<u> ⇵</u> 2s²<u>⇵</u> 2p₋₁<u>↑</u> 2p₀<u>↑</u> 2p₊₁<u>∅</u>
After Hybridization => C: 1s² 2sp¹ 2sp¹ 2p₀¹2p₊₁¹ => 2 hybrid orbitals and 2 unhybridized p-orbitals at the n=2 energy level.