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2 years ago

For Carbon, which sublevel has unpaired electrons? ASAP

1 answer:

7 0

When it says Carbon has 2 unpaired electrons, is referring to the fact that Carbon has an electron configuration of [He]2s22p2. Therefore Carbon has 2 unpaired electrons in the p sublevel.

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a 1.00-l solution contains 2.50×10-4 m cu(no3)2 and 1.20×10-3 m ethylenediamine (en). the kf for cu(en)22 is 1.00 × 1020. chegg

Debora [2.8K]

The concentration of copper in the solution is

$5.1 \times 10^{18} \: M.$

The volume of the solution = 1.00 L

$Moles \: of \:Cu(NO_{3}) _{2} = 2.50 \times 10 {}^{ - 4}$

$Moles \: of \: Ethylenediamine = 1.20 \times 10.3$

$k _{f} = 1.00 \times 10^{20}$

Overall balanced equation of reaction is,

$Cu^{2 + }(aq) \: + 2en \: (aq)→ Cu(en) ^{2 + }_{2}(aq)$

$Moles \: of \:Cu(NO_{3}) _{2} =0.00025 \: mol$

Mole ratio for,

$Moles \: of\:Cu(NO_{3}) _{2}: en$

$2: 1$

$Moles \: of \: en = 0.000250 \times 2$

$= 0.00050 \: moles$

Remained moles of en are= (0.00120-0.000500)

$= 0.000700 \: moles$

The concentration of copper in the solution is,

$k_{f} = \frac{Cu(en)^{2 + }_{2} }{Cu^{2+}en^{2} }$

$1.00 \times 10 ^{20} = \frac{0.000250}{(Cu )^{2 +} \times (0.000700)^{2 + } }$

$= 5.1 \times 10^{18} \: M.$

Therefore, the concentration of copper in the solution is

$5.1 \times 10^{18} \: M.$

To know more about molarity, refer to the below link:

brainly.com/question/8732513

#SPJ4

4 0

1 year ago

Write the complete balanced equation for the reaction between aluminum metal (Al) and oxygen gas (O2).

Arlecino [84]

Your answer is sooo close!!! But I think it is precisely about the same. :)
4AI(s) + 302(g) = 2 AI203(s)

7 0

2 years ago

An atom has 23 protons and 29 neutrons. Which is the correct chemical symbol for this atom?

Gnoma [55]

The symbol for this element is <span>V </span>, which is the well-known element vanadium.

5 0

3 years ago

Read 2 more answers

A sample of lead has the density of 58 g/mL and a mass of 15.2g, what is the volume of the lead?

torisob [31]

Answer:

0.26 ml

Explanation:

d = m/V

=> V = m/d = 15.2/58 = 0.26 ml

4 0

1 year ago

Excess silver(I) nitrate was added to a 8.500 g mixture containing some amount of barium chloride, and 7.123 g of silver chlorid

dsp73

Answer:

60.88%

Explanation:

The balanced equation of the reaction is given as;

2 AgNO3 (aq) + BaCl2 (aq) → 2 AgCl (s) + Ba(NO3)2 (aq)

Since AgNO3 is in excess, the limiting reactant is BaCl2. From the reaction;

1 mol of BaCl2 produces 2 mol of AgCl

Converting to masses;

Mass = Number of mol * Molar mass

BaCl2;

Mass = 1 * 208.23 g/mol = 208.23 g

AgCl;

Mass = 2 * 143.32 g/mol = 286.64 g

208.23 g BaCl2 produces 286.64 g of AgCl

x g BaCl2 produces 7.123 g of AgCl

Solving for x;

x = 7.123 * 208.23 / 286.64 = 5.1745 g

Mass percent = Mass / Total mass of Mixture * 100

Mass Percent = 5.1745 / 8.500 = 0.6088 * 100 = 60.88%

8 0

2 years ago