Answer:
(a) Acid
(b) Base
(c) Acid
(d) Base
Explanation:
According to the Arrhenius acid-base theory:
- An acid is a substance that releases H⁺ in aqueous solution.
- A base is a substance that releases OH⁻ in aqueous solution.
(a) H₂SO₄ is an acid according to the following equation:
H₂SO₄(aq) ⇒ 2 H⁺(aq) + SO₄²⁻(aq)
(b) Sr(OH)₂ is a base according to the following equation:
Sr(OH)₂(aq) ⇄ Sr²⁺(aq) + 2 OH⁻(aq)
(c) HBr is an acid according to the following equation:
HBr(aq) ⇒ H⁺(aq) + Br⁻(aq)
(d) NaOH is a base according to the following equation:
NaOH(aq) ⇒ Na⁺(aq) + OH⁻(aq)
Answer:
About 0.0940 M.
Explanation:
Recall that NaOH is a strong base, so it dissociates completely into Na⁺ and OH⁻ ions. Because the acid is monoprotic, we can represent it with HA. Thus, the reaction between HA and NaOH is:
Using the fact that it took 15.00 mL of NaOH to reach the endpoint, determine the number of HA that was reacted with:
Therefore, the molarity of the original solution was:
![\displaystyle \left[ \text{HA}\right] = \frac{0.00188\text{ mol}}{20.00\text{ mL}} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 0.0940\text{ M}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cleft%5B%20%5Ctext%7BHA%7D%5Cright%5D%20%3D%20%5Cfrac%7B0.00188%5Ctext%7B%20mol%7D%7D%7B20.00%5Ctext%7B%20mL%7D%7D%20%5Ccdot%20%5Cfrac%7B1000%5Ctext%7B%20mL%7D%7D%7B1%5Ctext%7B%20L%7D%7D%20%3D%200.0940%5Ctext%7B%20M%7D)
In conclusion, the molarity of the unknown acid is about 0.0940 M.
Answer:
volume of the gas is 5.0L
Explanation:
Using Boyle's law that state the pressure of a gas is inversely proportional to volume of it occupies when temperature is constant, it is possible to write:
P₁V₁ = P₂V₂
<em>Where P is pressure, V is volume and 1 and 2 are initial and final states.</em>
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If initial volume is 2.5L, initial pressure is 2.0atm and 1.0atm is final pressure, final volume is:
2.0atm*2.5L = 1atm V₂
5.0L = V₂
Thus, <em>volume of the gas is 5.0L</em>.
I need the answers or a picture to help
Mass = molarity x molar mass( NaNO₃) x volume
mass = 1.50 x 85.00 x 4.50
mass = 573.75 g of NaNO₃
hope this helps!
