English please i cant understand
Answer:
5.00 grams of salt contain more particles than 5.0 grams of sugar
Explanation:
Salt = NaCl
Molar mass = 58.45 g/mol
Sugar = C₁₂H₂₂O₁₁
Molar mass = 342.3 g/mol
Sugar's molar mass is higher than salt.
So 1 mol of sugar weighs more than 1 mol of salt
But 5 grams of salt occupies more mole than 5 grams of sugar
5 grams of salt = 5g / 58.45 g/m = 0.085 moles
5 grams of sugar = 5g/ 342.3 g/m = 0.014 moles
In conclusion, we have more moles of salt in 5 grams; therefore there are more particles than in 5 g of sugar.
When `CO_(2)` is bubbled through a cold pasty solution of barium peroxide in water, `H_(2)O_(2)` is obtained. <br> `BaO+CO_(2)+H_(2)OtoBaCO_(3)+H_(2)O_(2)` Barium carbonate being insoluble is filtered off. This is known as Merck's process.
<h3>What is meant by Perhydrol?</h3>
perhydrol (countable and uncountable, plural perhydrols) A stabilised solution of hydrogen peroxide.
<h3>What is Merck's Perhydrol?</h3>
Uses: Perhydrol is used as an antiseptic for wounds, and also acts as a germicide to kill bacteria and germs.
Being a strong oxidizing agent it has bleaching properties and acts as a ripening agent.
Learn more about merck's process here:
<h3>
brainly.com/question/16856280</h3><h3 /><h3>#SPJ4</h3>
Answer:
The correct option is D
Explanation:
Normally, beta-oxidation of fatty acid occurs in the mitchondrial matrix, however, when the fatty acid chains are too long, the beta-oxidation occurs in the peroxisomes <u>where the oxidation is not attached to ATP synthesis but rather transferred (i.e high energy electrons are transferred) to O₂ to form hydrogen peroxide</u> (H₂O₂). This is the major difference between the beta-oxidation that occurs in the peroxisomes to that which occurs in the mitochondria.
Answer: Concentration of
in the equilibrium mixture is 0.31 M
Explanation:
Equilibrium concentration of
= 0.729 M
The given balanced equilibrium reaction is,

Initial conc. x 0 0
At eqm. conc. (x-2y) M (y) M (3y) M
The expression for equilibrium constant for this reaction will be:
3y = 0.729 M
y = 0.243 M
![K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5By%5D%5Ctimes%20%5B3y%5D%5E3%7D%7B%5Bx-2y%5D%5E2%7D)
Now put all the given values in this expression, we get :



concentration of
in the equilibrium mixture = 
Thus concentration of
in the equilibrium mixture is 0.31 M