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Nonamiya [84]
3 years ago
9

How many sub-energy levels does the 2nd MEL have?

Chemistry
1 answer:
DedPeter [7]3 years ago
5 0

Answer:

n=  | Shell            |       Maximum Number of Electrons

1    | 1st Shell      |      2

2   | 2nd Shell      |      8

3   | 3rd Shell        |      18

4   |  4th Shell        |      32

Explanation: cause :)

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Hich of the following statements is true about what happens during a chemical reaction?
postnew [5]
I believe it is <span>d. the bonds of both the reactants and the products are formed.</span>
6 0
3 years ago
What is the molality of a solution composed of 342 moles of sugar (C12H22O11) dissolved in 171 kilograms of water (H20)?
4vir4ik [10]
Molality=mol/kg

342/171

=2m

Hope this helped :)
5 0
1 year ago
I. Choose the letter of the correct answer. Write your answer on a separate sheet of paper.
omeli [17]

Answer:

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Explanation:

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6 0
2 years ago
determine the freezing point depression of a solution that contains 30.7 g glycerin (c3h8o3, molar mass
Blababa [14]

The freezing point depression of a solution containing 30.7 g of glycerin  is  calculated as -1.65°C

Equating :

It is given that,

Given mass of glycerin is = 30.7 grams (Solute)

Volume of water = 376 mL

K_{f}or molar -freezing-depression point is = 1.86°C/m

Molar mass of glycerin = 92.09 g/mole

Now, to work out the value, the mass of water should be known. Thus, to calculate, the formula used will be:

Mass = Density X Volume

Mass = 1.0 g/mL X 376 mL

Mass = 376 g or 0.376 Kg

Using the formula of melting point depression, the equation becomes:

             ΔT_{f} = i ×K_{f} ×m

T⁰-T_{s}  = i *K_{f} *\frac{mass of glycerin}{molar mass of glycerin * mass of water     in     kg}

in which,

ΔT_{f} = change in freezing point

ΔT_{s} = freezing point of solution that has to be find

ΔT° = freezing point of water ()

Since, glycerin is a non-electrolyte, the Van't Hoff factor will be 1.

Substituting the values in the above equation:

0⁰C₋T_{s} = 1 ×1.86°C/m ×\frac{30.7}{92.09g/mol * 0.376kg}

T_{s} = -1.65°C

Thus, the freezing point depression of a solution is  -1.65°C

<h2 />

Freezing point depression

Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all less than that of the pure solvent and is directly proportional to the molality of the solute

Is melting point elevation or depression?

Boiling point elevation is that the raising of a solvent's boiling point due to the addition of a solute. Similarly, melting point depression is the lowering of a solvent's freezing point due to the addition of a solute. In fact, because the boiling point of a solvent increases, its melting point decreases

Learn more about freezing point depression :

brainly.com/question/26525184

#SPJ4

8 0
10 months ago
The decomposition of N_2O_5(g) following 1st order kinetics. N_2O_5(g) to N_2O_4(g) + ½ O_2(g) If 2.56 mg of N_2O_5 is initially
Crank

Answer: The rate constant is 0.334s^{-1}

Explanation ;

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = age of sample  = 4.26 min

a =  initial amount of the reactant  = 2.56 mg

a - x = amount left after decay process  = 2.50 mg

Now put all the given values in above equation to calculate the rate constant ,we get

4.26=\frac{2.303}{k}\log\frac{2.56}{2.50}

k=\frac{2.303}{4.26}\log\frac{2.56}{2.50}

k=5.57\times 10^{-3}min^{-1}=5.57\times 10^{-3}\times 60s^{-1}=0.334s^{-1}

Thus rate constant is [tex]0.334s^{-1}

4 0
3 years ago
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