Answer:
9.6 moles O2
Explanation:
I'll assume it is 345 grams, not gratis, of water. Hydrogen's molar mass is 1.01, not 101.
The molar mass of water is 18.0 grams/mole.
Therefore: (345g)/(18.0 g/mole) = 19.17 or 19.2 moles water (3 sig figs).
The balanced equation states that: 2H20 ⇒ 2H2 +02
It promises that we'll get 1 mole of oxygen for every 2 moles of H2O, a molar ratio of 1/2.
get (1 mole O2/2 moles H2O)*(19.2 moles H2O) or 9.6 moles O2
Answer:
distilled water I guess !
Just a guess though
If you find this useful, please mark my answer as the brainliest.
Explanation:
If you find this useful, please mark my answer as the brainliest.
distilled water
Explanation:
Some students investigated osmosis in raw potato sticks. The students measured the mass of three potato sticks using an electronic balance. The students left each potato stick in one of the three different liquids for 5 hours:i. distilled water. ii. dilute sodium chloride solution. iii. concentrated sodium chloride solution. After 5 hours they measured the mass again and calculated the change in mass. 1. Predict which of the liquids would cause the largest decrease in mass of a potato stick. 2. After the experiment, the students noticed that the potato stick with the lowest mass was soft and floppy. Explain why the potato stick had become soft and floppy. 3. The students followed the same experimental procedure with boiled potato sticks and found no overall change in mass in any of the solutions. Suggest why the mass of the boiled potato sticks remained the same.
Explanation:
The given precipitation reaction will be as follows.
Here, AgCl is the precipitate which is formed.
It is known that molarity is the number of moles present in a liter of solution.
Mathematically, Molarity = 
It is given that volume is 1.14 L and molarity is 0.269 M. Therefore, calculate number of moles as follows.
Molarity =
0.269 M = 
no. of moles = 0.306 mol
As molar mass of AgCl is 143.32 g/mol. Also, relation between number of moles and mass is as follows.
No. of moles = 
0.307 mol = 
mass = 43.99 g
Thus, we can conclude that mass of precipitate produced is 43.99 g.
<span>The average molar bond enthalpy of the carbon-hydrogen bond in a CH4 molecule is 416 KJ/mol.
(+716.7 + (4 x 218) - (- 74.6) ) / 4
= + 1663.3 / 4
= 416</span>
