A Bronsted-Lowry acid is a chemical species that donates one or more hydrogen ions in a reaction. In contrast, a Bronsted-Lowry base accepts hydrogen ions. When it donates its proton, the acid becomes its conjugate base. A more general look at the theory is as an acid as a proton donor and a base as a proton acceptor. :)
Answer:
False
Explanation:
Molecules are not more sizable atoms than the average atom. Molecules are compunds of two different symbolic elements, when you combine then you get a molecule. The answer to your question is false because if they were larger molecules, they would be in object around us but molecules are not in all objects around us.
The major classes are monosaccharides, oligosaccharides and polysaccharides.
The types are sugar, starch, and fibersugar.
Hope I helped!!
Answer: a) 
b) 
Explanation:
The reaction is :
Rate = Rate of disappearance of 
= Rate of appearance of 
Rate = ![-\frac{d[N_2O_5]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BN_2O_5%5D%7D%7B2dt%7D)
= ![\frac{d[NO_2]}{4dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO_2%5D%7D%7B4dt%7D)
Rate of disappearance of = 
= 
a) Rate of disappearance of =
Rate of appearance of =
b) Rate of appearance of = ![\frac{d[NO_2]}{dt}=2\times 1.7\times 10^{-4}}=3.4\times 10^{-4}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D2%5Ctimes%201.7%5Ctimes%2010%5E%7B-4%7D%7D%3D3.4%5Ctimes%2010%5E%7B-4%7D)
Answer:
1.62 × 10²⁴ atoms are in 52.3 g of lithium hypochlorite.
Explanation:
To find the amount of atoms that are in 52.3 g of lithium hypochlorite, we must first find the amount of moles. We do this by dividing by the molar mass of lithium hypochlorite.
52.3 g ÷ 58.4 g/mol = 0.896 mol
Next we must find the amount of formula units, we do this be multiplying by Avagadro's number.
0.896 mol × 6.02 × 10²³ = 5.39 × 10²³ f.u.
Now to get the amount of atoms we can multiply the amount of formula units by the amout of atoms in one formula unit.
5.39 × 10²³ f.u. × 3 atom/f.u. = 1.62 × 10²⁴ atoms
1.62 × 10²⁴ atoms are in 52.3 g of lithium hypochlorite.
