Temperature and salinity changes the density of water.
<h3>Effect of Temperature and salinity on water</h3>
Temperature and salinity directly affect density of water. Water that low temperature is more denser than water that has high temperature while on the other hand, Freshwater which has no salt is less denser than seawater which has more salt concentration so we can conclude that temperature and salinity changes the density of water.
Learn more about salinity here: brainly.com/question/20283396
Is there something else that goes along with the question?
Answer:
Explanation:
<u>1) Data:</u>
Base: NaOH
Vb = 15.00 ml = 15.00 / 1,000 liter
Mb = ?
Acid: H₂SO₄
Va = 17.88 ml = 17.88 / 1,000 liter
Ma = 0.1053
<u>2) Chemical reaction:</u>
The <em>titration</em> is an acid-base (neutralization) reaction to yield a salt and water:
- Acid + Base → Salt + Water
- H₂SO₄ (aq) + NaOH(aq) → Na₂SO₄ (aq) + H₂O (l)
<u>3) Balanced chemical equation:</u>
- H₂SO₄ (aq) + 2 NaOH(aq) → Na₂SO₄ (aq) + 2H₂O (l)
Placing coefficient 2 in front of NaOH and H₂O balances the equation
<u>4) Stoichiometric mole ratio:</u>
The coefficients of the balanced chemical equation show that 1 mole of H₂SO₄ react with 2 moles of NaOH. Hence, the mole ratio is:
- 1 mole H₂SO₄ : 2 mole NaOH
<u>5) Calculations:</u>
a) Molarity formula: M = n / V (in liter)
⇒ n = M × V
b) Nunber of moles of acid:
- nₐ = Ma × Va = 0.1053 (17.88 / 1,000)
c) Number of moles of base, nb:
- nb = Mb × Vb = Mb × (15.00 / 1,000)
d) At equivalence point number of moles of acid = number of moles of base
- 0.1053 × (17.88 / 1,000) = Mb × (15.00 / 1,000)
- Mb = 0.1053 × 17.88 / 15.00 = 0.1255 mole/liter = 0.1255 M
Answer:
4 1/2
Explanation:
Use a ratio to find your answer
4 6
----- = -------
3 x
Cross multiply to solve for x.
4x = 18
x = 18/4
x = 4 2/4 which is the same as 4 1/2
