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For each of the following sublevels, give the n and l values and the number of orbitals: (a) 5s; (b) 3p; (c) 4f

OverLord2011 [107]

Answer:

(a) 5s. n = 5. Sublevel s, l = 0. Number of orbitals = 1

(b) 3p. n = 3. Sublevel p, l = 1. Number of orbitals = 3

(c) 4f. n =4. Sublevel f, l = 3. Number of orbitals = 7

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n

2. Sublevel number, 0 ≤ l ≤ n − 1

So,

(a) 5s. n = 5, shell number 5. Sublevel s, l = 0. Number of orbitals = 2l +1 = 1

(b) 3p. n = 3, shell number 3. Sublevel p, l = 1. Number of orbitals = 2l +1 = 3

(c) 4f. n =4, shell number 4. Sublevel f, l = 3. Number of orbitals = 2l +1 = 7

6 0

3 years ago

A chemistry student weighs out of sulfurous acid , a diprotic acid, into a volumetric flask and dilutes to the mark with distill

nadezda [96]

The question is incomplete, here is the complete question:

A chemistry student weighs out 0.104 g of sulfurous acid, a diprotic acid, into a 250.0 mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.

<u>Answer:</u> The volume of NaOH needed is 36.2 mL

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of sulfurous acid = 0.104 g

Molar mass of sulfurous acid = 82 g/mol

Volume of solution = 250 mL

Putting values in above equation, we get:

$\text{Molarity of sulfurous acid}=\frac{0.104\times 1000}{82\times 250}\\\\\text{Molarity of sulfurous acid}=5.07\times 10^{-3}M$

To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=5.07\times 10^{-3}M\\V_1=250mL\\n_2=1\\M_2=0.0700M\\V_2=?mL$

Putting values in above equation, we get:

$2\times 5.07\times 10^{-3}\times 250.0=1\times 0.0700\times V_2\\\\V_2=\frac{2\times 5.07\times 10^{-3}\times 250}{1\times 0.0700}=36.2mL$

Hence, the volume of NaOH needed is 36.2 mL

5 0

3 years ago

Nitrogen monoxide, NO, reacts with hydrogen, H₂, according to the following equation.

Anvisha [2.4K]

Answer : The rate law for the overall reaction is, $Rate=k[NO]^2[H_2]$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

As we are given the mechanism for the reaction :

Step 1 : $2NO+H_2\rightarrow N_2+H_2O_2$ (slow)

Step 2 : $H_2O_2+H_2\rightarrow 2H_2O$ (fast)

Overall reaction : $2NO+2H_2\rightarrow N_2+2H_2O$

The rate law expression for overall reaction should be in terms of $NO\text{ and }H_2$ .

As we know that the slow step is the rate determining step. So,

The slow step reaction is,

$2NO+H_2\rightarrow N_2+H_2O_2$

The expression of rate law for this reaction will be,

$Rate=k[NO]^2[H_2]$

Hence, the rate law for the overall reaction is, $Rate=k[NO]^2[H_2]$

8 0

4 years ago

The temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 47 calories of heat.

sleet_krkn [62]

This would be concidered a chemical reaction.

6 0

4 years ago

PLS HELP NOW WILL GIVE BRAINLIEST

zvonat [6]

Answer:

Liquids, because they flow, can occupy whatever shape their container has, so they do not have a fixed shape. Because the particles in liquids are very close together (barely further apart than in solids) liquids do not easily compress, so their volume is fixed. hope that helps love!

7 0

3 years ago

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