Explanation:
1.
Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)
2.
Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)
A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.
3.
Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)
2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)
Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)
2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)
4.
The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.
Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.
To find for the oxidizing agent, first let us write the
half reactions of this complete chemical reaction:
Ca = Ca2+ + 2e- <span>
2 H+ + 2e- = H2</span>
The oxidizing agent
would be the substance of the element that is reduced. We know that an element
is reduced when an electron is added to it. In this case, the element being
reduced is H. Therefore the oxidizing agent is HNO3.
Answer:
<span>HNO3</span>
Explanation:
Moles of NaOH = 10g / (40g/mol) = 0.25mol.
0.25mol / 500g = 0.50mol / 1000g = 0.50mol/dm³.
The molarity is 0.50mol/dm³.
Answer:
Formula: Na2S2O3
we get solubility.
Divide the mass of the compound by the mass of the solvent and then multiply by 100 g to calculate the solubility in g/100g .
Solution given:
mass of sodium thiosulphate [m1]=25.5g
mass of water [m2]=40g
at temperature [t]=25°C
we have
<u>solubility in g/dm^3</u> :
- =

- =63.75g /litre=63.75g/dm³
<u>solubility in g/dm^3 :63.75g/dm³</u>
<u>n</u><u>o</u><u>w</u>
solubility of the solute in mol/dm^3=:63.75g/dm³/178=0.4 mol/dm³
Answer:
The answer is both molecule and an element
Hydrogen is the chemical element with the symbol H and atomic number 1. ... At standard conditions hydrogen is a gas of diatomic molecules having the formula H2. It is colorless, odorless, non-toxic, and highly combustible.
Explanation: