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Sunny_sXe [5.5K]

3 years ago

8

How many equivalent resonance forms can be drawn for co32-? (carbon is the central atom.) how many equivalent resonance forms ca

n be drawn for co32-? (carbon is the central atom.) 4 2 1 0 3?

1 answer:

mylen [45]3 years ago

3 0

Check the attached file for the solution.

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Federal regulations set an upper limit of 50 parts per million (ppm) of NH3 in the air in a work environment [that is, 50 molecu

pogonyaev

Answer:

1) A total of 5.91×10-3 grams were drawn into the HCl solution

2) 84.3 ppm of NH3 were in the air

3) As 84.3 ppm is higher than 50 ppm established in the regulation, the manufacturer does not comply with it.

Explanation:

The problem shows the following process: an amount of air with NH3 passes through a HCl solution. The NH3 reacts with HCl reducing the concentration of the latted and finally the remaining HCl is titrated with NaOH.

The process to solve this problem should go as follows:

a) Calculate the amount of the remaining HCl that was titrated with the NaOH at the end.

b) Calculate the amount of HCl that reacted with NH3, using the data from a)

c) Calculate the amount of NH3 present in air using the data from b)

d) Calculate the grams of NH3 using the data from c) to solve question 1)

e) Calculate the number of moles of air

f) Calculate the ppm of NH3 in air using the data from c) and e) to solve questions 2) and 3)

So, let's proceed:

a) To do this we need to take a look at the chemical equation of the HCl and NaOH reaction:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

And we see that 1 mole of HCl reacts with 1 mole of NaOH. Therefore, being n the number of moles:

n(NaOH) = n(HCl)

(5.86×10-2 M)×(14.5×10-3 L) = n(HCl)

n(HCl) = 8.50×10-4 moles of HCl

b) So, as we now have the amount of the remaining HCl, we need to find out how much HCl reacted with NH3 in the first place, so we need to substract the number of moles found in a) from the number of moles we had initially:

n(HCl)_reacted with NH3 = n(HCl)_initial - n(HCl)

n(HCl)_reacted with NH3 = (1.13×10-2 M)×(106×10-3 L) - 8.50×10-4

n(HCl)_reacted with NH3 = 3.49×10-4 moles of HCl

c) Now, as we know the amount of HCl that reacted with NH3, we can calculate the amount of NH3 that was drawn into the solution using the chemical equation (fortunately, the equation is already balanced):

NH3(aq) + HCl(aq) → NH4Cl(aq)

And we can see 1 mole of NH3 reacts with 1 mole of HCl, so we can conclude that 3.49×10-4 moles of HCl have reacted with 3.49×10-4 moles of NH3.

As we are being asked by the grams, we must convert that using the molar mass of NH3 that is 17 g/mol (N=14, H=1), so:

grams of NH3 = (3.49×10-4 mol)×(17 g/mol) = 5.91×10-3 grams of NH3

d) Now we must calculate the number of moles of air in order to be able to calculate the parts per million of NH3:

In this case we have to notice that we have passes air at a rate of 10.0 liters per minute and we have done it by 10 minutes, that means that the total amount of air (in liters) we have passed through the solution is:

liters of air = 10 min × 10 L/min = 100 L

e) That volume of air can be converted into moles using the information from question 2):

moles of air = (100 L) × (1.2 g/L) × (1 mol/29 g) = 4.14 moles

f) We can calculate now using the information from c) and e) as follows:

ppm of NH3 in air = number of moles of NH3 / number of moles of air × 1000000

ppm of NH3 = (3.49×10-4 mol)/(4.14 mol)×1000000 = 84.3 ppm

In conclusion, the manufacturer does not comply with the regulation of maximum 50ppm of NH3.

6 0

4 years ago

what is the period of oscillation of an electron that is bouncing up and down in response to the passage of a packet of red lgih

Talja [164]

Answer:

2.33*10^-15 s

Explanation:

We need to estimate the period of oscillation of an electron that bounced up/down due to the effects of a passing packet of red light. The period of oscillation is the wavelength of the red light divided by the speed of the red light. We know that:

V = f*λ

where V = speed of the red light = 3*10^8 m/s

f = frequency = 1/T where T is the period of oscillation in seconds (s)

λ = the wavelength of the red light = 0.7*10^-6 m

Thus:

V = λ/T

T = λ/V = (0.7*10^-6)/(3*10^8) = 2.33*10^-15 s

7 0

3 years ago

Consider the reaction between hcl and o2: 4hcl(g)+o2(g)→2h2o(l)+2cl2(g) when 63.1 g of hcl are allowed to react with 17.2 g of o

Roman55 [17]

The balanced chemical reaction is expressed as:

<span>4hcl(g)+o2(g)→2h2o(l)+2cl2(g)

To determine the percent yield of the reaction, we need to calculate for the theoretical yield. This is the maximum amount of the product that can be produced from the reaction given the initial amounts of the reactants. First, we identify the limiting reactant as follows:

</span><span>63.1 g of hcl ( 1 mol / 36.46 g ) ( 1 mol O2 / 4 mol HCl) ( 32 g / mol) = 13.85 g O2
17.2 g of o2 ( 1 mol / 32 g ) ( 4 mol HCl / 1 mol O2) ( 36.46 g / mol) = 78.39 g HCl

Therefore, the limiting reactant would be HCl. We use the value for the HCl to calculate for the theoretical yield.

</span>63.1 g of hcl ( 1 mol / 36.46 g ) ( 2 mol Cl2 / 4 mol HCl) ( 70.9 g / mol ) = 61.35 g Cl2

Percent yield = actual / theoretical x 100
= 59.6 / 61.35 x 100
= 97.1%

4 0

3 years ago

Which explanation describes the last part of a chemical reaction?

Cloud [144]

Answer:

Transition state falls apart to form products.

Explanation:

Reactants collide to form a transition state is actually the initial step (not the last step) because reactants have react with each other. Transition state is the middle stage, once this stage falls or passes, you get products. I have attached an example reaction scheme to better understand.

8 0

2 years ago

0.55 moles of sodium nitrate to molecules of sodium nitrate.

scoundrel [369]

<h3>Answer:</h3>

3.31 × 10^23 molecules

<h3>Explanation:</h3>

We are given;

The number of moles = 0.55 moles

We are required to calculate the number of molecules in 0.55 moles of sodium nitrate.

We know that 1 mole of a compound contains molecules equivalent to the Avogadro's number, 6.022 × 10^23.

Therefore;

1 mole of NaNO₃ = 6.022 × 10^23 molecules

Thus, for 0.55 moles will contain;

= 0.55 moles × 6.022 × 10^23 molecules

= 3.31 × 10^23 molecules

Hence, 0.55 moles will contain 3.31 × 10^23 molecules

4 0

3 years ago