X moles of CO₂, and x moles of H₂O
CxH2x + 3x/2 O₂ -------> xCO₂ + xH₂O
CxH2x or CnH2n ----> ALKENE ( hydrocarbon with double bond)
Answer:
The soil has a percentage of water by mass of 27.8 %.
Explanation:
Keeping in mind that
- Mass of Clay = Mass of water + Mass of Dry Soil
we can <u>calculate the mass of water</u>:
- 1350 g = Mass of Water + 975 g
We can then <u>calculate the mass percentage of water in the soil</u>:
- 375 / 1350 * 100% = 27.8 %
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
I think the answer would be the number of electron in pairs that is around the central atom. This is the most general principle in determining structure of a molecule. Electrons are the ones responsible in bonding with other atoms. Hope this helped.
The concentration of diluted solution is found to be 0.421 M
<u>Explanation:</u>
It is given that a stock solution (400 ml) should be diluted to 750 ml, we need to find the concentration of the diluted solution using the Law of Volumetric analysis,
V1M1 = V2M2
V1 being the volume of the stock solution = 400 ml
M1 being the molarity of the stock solution = 0.789 M
V2 being the volume of the diluted solution = 750 ml
M2 being the molarity of the diluted solution = ?
Now we have the molarity of the diluted solution by rearranging the above equation as,
M2 = 
Now plugin the values as,
M2 = 
= 0.421 M
So the molarity of diluted solution is found to be 0.421 M.
