★ « <em><u>what is oxidation number of S in H2SO5??</u></em><em><u> </u></em><em><u>»</u></em><em><u> </u></em><em><u>★</u></em>
- <em><u>it's </u></em><em><u> </u></em><em><u>6</u></em><em><u>!</u></em><em><u>!</u></em>
Explanation:
- <em>Oxidation number of S in H2SO5 is 6 .</em>
Answer:(4) ----accepts a proton
Explanation:
H2O water can produce both hydrogen and hydroxide ions
H2O --> H+ + OH-
According to the Bronsted-Lowry theory, it can be a proton donor and a proton acceptor.this means that It can donate a hydrogen ion to become its conjugate base, or can accept a hydrogen ion to form its conjugate acid,
When , a water molecule, H2O accepts a proton it will act as a Brønsted-Lowry base especially when dissolved in a strong acidic medium. for eg
HCl + H2O(l) → H3O+(aq) + Cl−(aq)
Here, Hydrochloric acid is a strong acid and ionizes completely in water, since it is more acidic than water, the water will act as a base.
Lysosome is B. breaks down waste materials
Vacuole is A. temporary storage
Chloroplast is C. converts energy
Answer:
V KOH = 41 mL
Explanation:
for neutralization:
- ( V×<em>C </em>)acid = ( V×<em>C </em>)base
∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L
∴ V H2SO4 = 41 mL = 0.041 L
∴ <em>C</em> KOH = 0.0050 N = 0.0050 eq-g/L
∴ E KOH = 1 eq-g/mol
⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L
⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH
⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)
⇒ V KOH = 0.041 L
Answer:
Explanation:
Hello,
In this case, the combustion of methane is shown below:
And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:
Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:
Best regards.
