6.4 * 6.02 * 10^23 = 3.8528*10^24 atoms
Don't let the fact that it's vanadium throw you off, avagadros constant stays the same for all elements
Answer:
Explanation:
NH3 + H3O+ --> NH4+ + H2O
equllibrium constant =K = [ H2O] [NH4+] / [NH3] [H3O+ ]
=
by inserting thier respecive values can you calcaulte, by the way coniseder [ H2O] =1 ,
<u>Answer:</u> It repels positive ion and attracts negative ion.
<u>Explanation:</u>
There are 2 types of ions:
1. Cations: These ions are positively charged ions which are formed when a substance looses electrons.
2. Anions: These ions are negatively charged ions which are formed when a substance gains electrons.
It is known that, like charges repel each other and unlike charges attract each other.
As, it is given that the substance is positively charged, so it will attract anion and repel cation.
Answer:
Explanation:
A1. Chemical indicator, any substance that gives a visible sign, usually by a colour change, of the presence or absence of a threshold concentration of a chemical species, such as an acid or an alkali in a solution. An example is the substance called methyl yellow, which imparts a yellow colour to an alkaline solution.
A2. The reaction of an acid with a base is called a neutralization reaction. The products of this reaction are a salt and water. ... For example, the reaction of hydrochloric acid, HCl, with sodium hydroxide, NaOH, solutions produces a solution of sodium chloride, NaCl, and some additional water molecules
A3. Methyl orange has the property to color alkaline and neutral water yellow. If the water becomes acidic, it turns red immediately.
Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
