1. Option A. Temperature and Salinity
2. Option D. (I think)
3. Option D.
Hope your test goes well!
A)Constant pumping from the heart which provides adequate pressure necessary to move the blood through the arteries and veins
B)Gravitational force
Answer:
The answer to your question: 0.7 M
Explanation:
Data
V of KOH = 90 ml
[KOH] = ?
V H2SO4 = 21.2 ml
[H2SO4] = 1.5 M
2KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2H₂O(l)
Molarity = moles / volume
moles of H₂SO₄ = (1.5) (21.2)
= 31.8
2 moles of KOH -------------- 1 mol of H₂SO₄
x -------------- 31.8 mol of H₂SO₄
x = (31.8)(2) / 1
x = 63.8 moles of KOH
Molarity = 63.8 / 90
= 0.7 M
Answer:
- <em>To balance a chemical equation it may be necessary to adjust the </em><u>coefficients.</u>
Explanation:
The <em>coefficients</em> of a <em>chemical equation</em> are the numbers that you put in front of each reactant and product. They are used to balance the equation and comply with the law of mass conservation.
By adjusting the coefficients you obtain the relative amounts (moles) of each product and reactant, i.e. the mole ratios.
Here an example.
The first information is what is called a word equation. E.g. nitrogen and hydrogen react to form ammonia:
- Word equation: hydrogen + nitrogen → ammonia
- Skeleton equation: H₂ + N₂ → NH₃
This equation shows the chemical formulae but it is not balanced. The law of mass conservation is not observed.
So, in order to comply with the law of mass conservation you adjust the coefficients as follow.
- Balanced chemical equation: 3H₂ + N₂ → 2NH₃
As you see, it was necessary to modify the coefficients. Now the law of conservation of mass is observed and you get the mole ratios:
- 3 mol H₂ : 1 mol N₂ : 2 mol NH₃
Answer :
Example of polar covalent molecules H-O-H(water), ammonia
Explanation:
The presence of intermolecular Hydrogen bonding makes the boiling point of water unexpectedly high, and the polar covalent nature makes it dissolve polar solute/compound
