1. What is the difference between charging by conduction and charging by induction?

1. If a solution containing 48.99 g of mercury(II) perchlorate is allowed to react completely with a solution containing 8.564 g

Vitek1552 [10]

1. Chemical eqn:

Hg(ClO4)2 + Na2S -> HgS + 2NaClO4

mercury perchlorate has a molar mass of 399.5g/mol (1d.p) and for Na2S it is 78.0g/mol (1d.p.)

no. of moles of mercury perchlorate= 48.99÷399.5= 0.12263mol(5s.f.)

no. of moles of Na2S= 8.564÷78.0= 0.10979mol ( 5s.f.)

so no. of moles of Hg(ClO4)2/ no. of moles of Na2S= 1/1 according to the eqn

so no. of moles of Hg(ClO4)2 needed if all Na2S is used up is 1/1×0.10979= 0.10979 mols

since no. of moles of mercury perchlorate needed < no. of moles of it provided, it is in excess and Na2S is the limiting factor.

since HgS is a solid and NaClO4 is aqueous, solid ppt formed will only be HgS.

no. of moles of HgS/ no. of moles of NaClO4= 1/1

no. of moles of HgS produced= 0.10979mols

molar mass of HgS= 232.7g/mol 1d.p.

grams of solid produced= 232.7×0.10979= 25.5g (3s.f.)

2. reactant in excess is Hg(ClO4)2,

no. of excess moles= 0.12263-0.10979= 0.01284mols

grams of excess reactant= 0.01284×399.5= 5.13g (3s.f.)

3 0

3 years ago

What is the reasoning for the location of sodium in the periodic table

tatyana61 [14]

Due to it's electronic configuration w/c is 1s2 2s2 2ps 3s1 considering the last w/c is 3s1, sodium should be in row 3 period a1.

3 0

3 years ago

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Which of the following metals is most reactive?<br> a. Mg<br> b. Rb<br> c. Cs<br> d. Li

natita [175]

The answer would be c :)

5 0

3 years ago

In the reaction between the strong acid HCI and the strong base NaOH which of the ions listed below is a spectator ion?

qwelly [4]

Answer:

Na+

Explanation:

The equation would be:

HCl (aq) + NaOH (aq) --> HOH (l) + NaCl (aq)

The equation is already balanced and the NaCl will disassociate in Na+ and Cl- and HCl will disassociate into H+ and Cl- and NaOH will disassociate into Na+ and OH-. Na+ is on both sides of the equation and stays the same, so Na+ will be the spectator ion.

3 0

3 years ago

Read 2 more answers

0.030 moles of a weak acid, HA, was dissolved in 2.0 L of water to form a solution. At equilibrium, the concentration of HA was

nikitadnepr [17]

Answer: The value of $k_{a}$ for the weak acid is $3.07 \times 10^{-4}$ .