Answer:
What is the molarity of a solution containing 5.00 moles of kcl in 2.00L of solution? Molarity= moles of solute/volume of solution in litre , so the problem looks like this : 7/. 569 , which is equivalent to 12.302 M .
Answer:
atoms of hydrogen are there in
35.0 grams of hydrogen gas.
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number
of particles.
To calculate the moles, we use the equation:
1 mole of hydrogen
=
atoms
17.5 mole of hydrogen
=
atoms
There are
atoms of hydrogen are there in
35.0 grams of hydrogen gas.
True
Explanation:
It is correct to say that the biosphere includes all of the fish that are in the ocean. Even though the ocean is part of the hydrosphere.
The biosphere includes all life form on earth wherever they may dwell from deep abyss to the high atmosphere.
The different spheres on earth are interacting with one another and they depend on each other.
Just like the earth systems, the spheres of the earth relates with each other. Even the ocean is contained in the geosphere.
learn more:
Biodiversity brainly.com/question/10723602
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Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.


There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
![[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6