Explanation:
“The isomers butane and methyl propane have the same molecular formula and different properties”, this is because structural isomers usually have different properties to their parent.
The answer is b. 4.2 mole. The balanced reaction formula is 2LiOH + H2SO4 -->Li2SO4 + 2H2O. And the ratio of mole number of the reactants is the same as the ratio of coefficients.
Answer: Uranium enrichment. Uranium is used to fuel nuclear reactors; however, uranium must be enriched before it can be used as fuel. Enriching uranium increases the amount of uranium-235 (U235) that can sustain the nuclear reaction needed to release energy and produce electricity at a nuclear power plant.
Answer:
The pH of the solution is 11.48.
Explanation:
The reaction between NaOH and HCl is:
NaOH + HCl → H₂O + NaCl
From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:
Now, we need to find the concentration of the OH⁻ ions.
![[OH^{-}] = \frac{n_{NaOH}}{V}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20)
Where V is the volume of the solution = 1.00 L
![[OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20%3D%20%5Cfrac%7B3.01%20%5Ccdot%2010%5E%7B-3%7D%20moles%7D%7B1.00%20L%7D%20%3D%203.01%20%5Ccdot%2010%5E%7B-3%7D%20mol%2FL%20)
Finally, we can calculate the pH of the solution as follows:
![pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52](https://tex.z-dn.net/?f=%20pOH%20%3D%20-log%28%5BOH%5E%7B-%7D%5D%29%20%3D%20-log%283.01%20%5Ccdot%2010%5E%7B-3%7D%29%20%3D%202.52%20)
Therefore, the pH of the solution is 11.48.
I hope it helps you!
Answer:
you must add 50 mL
Explanation:
Hi
KOH is a strong base and by adding 100mL 0.05M you will have an amount of 5 millimol.
NaCN is a base and by adding 50 mL 0,150 M you will have an amount of 7,5 mmol.
HCl is a acid and by adding 200 mL 0,075 M you will have an amount of 15 mmol.
The acid reacts with the bases leaving 2.5 mmol unreacted.
Na3PO4 is a base and by adding 50 mL 0,1 M you will have an amount of 5 mmol.
The 2.5 mmol of acid react with the base PO4 ^ -3 forming a regulatory solution of PO4 ^ -3 and HPO4 ^ -2 of pKa 2.12
5 mmol of acid (HNO3) must be added to obtain a regulatory solution formed by the same amount of HPO4 ^ -2 (2.5 mmol) and H2PO4 ^ -1 (2.5 mmol) with pKa 7.21
Considering a quantity of 5 mmol of HNO3 of concentration 0.1 M, 50 mL must be added.
To calculate the pH of the regulatory solution you should consider pH = pKa × Ca / Cb pH = 7.21 × 2.5 / 2.5 = 7.21 Being in the same solution the volume is the same and can be simplified to achieve a faster calculation.
successes with your homework
