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3 years ago

5

In the Bohr model of the atom, an electron in an orbit has a fixed energy level and to move between orbitals what needs to absor

bed or released?
quantum of energy
light
heat
electricity

1 answer:

soldier1979 [14.2K]3 years ago

3 0

The answer is heat hope i helped

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Put the sets of 3 elements in order from least metallic character to most metallic character. Help please!!

9966 [12]

Answer:

29:Mn,V,Sr

30:Ni,Pd,Cs

31:Cr, Mo, W

32:Sn,Pb,Ti

33:F, P, As

Sorry I was in a rush so it may not be right; so check with this picture to help you out.

5 0

3 years ago

How many moles of water will be produced from the combustion of 0.27 moles of CH3OH?

Valentin [98]

Answer:

0.54 mole of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2CH3OH + 3O2 —> 2CO2 + 4H2O

From the balanced equation above,

2 moles of CH3OH reacted to produce 4 moles of water.

Finally, we shall determine the number of mole of water (H2O) produced by the reaction of 0.27 moles of CH3OH. This can be obtained as follow:

From the balanced equation above,

2 moles of CH3OH reacted to produce 4 moles of water.

Therefore, 0.27 moles of CH3OH will react to produce = (0.27 × 4)/2 = 0.54 mole of H2O.

Thus, 0.54 mole of H2O is produced from the reaction.

5 0

3 years ago

A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction

Molodets [167]

<u>Answer:</u> The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

<u>For NaOH:</u>

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol$

<u>For sulfuric acid:</u>

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, $2.925\times 10^{-3}$ moles of KOH will react with = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = $(4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol$

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, $2.925\times 10^{-3}$ moles of KOH will produce = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sodium sulfate

<u>For sodium sulfate:</u>

Moles of sodium sulfate = $1.462\times 10^{-3}moles$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M$

<u>For sulfuric acid:</u>

Moles of excess sulfuric acid = $3.498\times 10^{-3}mol$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M$

<u>For NaOH:</u>

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0}{0.050}=0M$

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

8 0

4 years ago

How many significant figures are in the number 020.310

True [87]

Answer:

5

Explanation:

Sorry, don't have one

<u><em>Hope this helps and I get brainliest <3</em></u>

6 0

3 years ago

If it requires 75.0mL of 0.500M NaOH to neutralize 165.0 mL of an hcl solution what is the concentration of the hcl solution

Kaylis [27]

Answer:

0.027 M HCl

Explanation:

The chemical equation of the neutralization is:

1 NaOH + 1 HCl -> 1 H2O + 1 NaCl

Because the ratio of NaOH and HCl is 1:1 you can use the M1V1=M2V2 formula.

(75 mL)(0.5 M NaOH) = (165 mL)(M HCl)

It requires 0.027 M HCl.

7 0

3 years ago