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Juli2301 [7.4K]
3 years ago
10

"what is the probability of finding an electron within one bohr radius of the nucleus"

Chemistry
1 answer:
Katena32 [7]3 years ago
6 0
Your answer is going to be b
You might be interested in
Experiment #1A melting point of an old sample of Naphthalene was completed and a melting range of 77-83 oC was observed and reco
White raven [17]

Answer:

%error = 0.32%

Explanation:

Let's answer both questions, by parts.

1. Percentage error:

In this case, I do not have the video, but I do have the reported melting point of naphtalene which is 80.26 °C.

The expression to calculate the percentage error is the following:

%Error = absolute error / actual percentage. (1)

And the absolute error is:

Abs error = actual value - experimental value  (2)

But the experimental value is a range, so we just have to get a average of that:

Exp value = 77 + 83 / 2 = 80 °C

Now the absolute error:

Abs error = 80.26 - 80 = 0.26 °C

Finally the %error:

%error = (0.26 / 80.26) * 100

<h2>%error = 0.32%</h2>

2. Meaning of melting point range and %error

The melting point range just means that the sample of naphtalene has impurities, and when a sample of any compound has impurities, melting point tends to be low. However, this decrease of temperature is a wider range. But usually a range of just 5° C means that compound has little traces of impurities but it can still be used for reactions.

The %error means that the impurities of the sample are really low, so the sample is practically pure with little traces of impurities.

3 0
3 years ago
If 2.0 ml of 6.0m hcl is used to make a 500.0-ml aqueous solution, what is the molarity of the dilute solution?
svp [43]
Make sure that you understand what they are asking you from this question, as it can be confusing, but the solution is quite simple. They are stating that they want you to calculate the final concentration of 6.0M HCl once a dilution has been made from 2.0 mL to 500.0 mL. They have given us three values, the initial concentration, initial volume and the final volume. So, we are able to employ the following equation:

C1V1 = C2V2
(6.0M)(2.0mL) = C2(500.0mL)
Therefore, the final concentration, C2 = 0.024M.
4 0
3 years ago
Read 2 more answers
Photosynthesis allows plants to turn light energy into chemical energy by forming glucose from carbon dioxide and water: 6CO2(g)
dezoksy [38]

Answer:

305 g of CO₂

3.77 × 10⁵ kJ

Explanation:

Let's consider the global reaction for photosynthesis.

6 CO₂(g) + 6 H₂O(l) → C₆H₁₂O₆(g) + 6 O₂(g)  ΔHrxn = 2802.8 kJ

<em>A 1.70 lb sweet potato is approximately 73% water by mass. If the remaining mass is made up of carbohydrates derived from glucose (MW = 180.156 g/mol), how much carbon dioxide (MW = 44.01 g/mol) was needed to grow this sweet potato?</em>

<em />

Let's consider the following relations:

  • The potato is 100%-73%=27% glucose by mass.
  • 1 lb = 453.59 g.
  • 6 moles of CO₂ produce 1 mole of glucose.
  • The molar mass of glucose is 180.156 g/mol.
  • The molar mass of carbon dioxide is 44.01 g/mol.

Then, for a 1.70 lb potato:

1.70lbPotato.\frac{27lbGlucose}{100lbPotato} .\frac{453.59gGlucose}{1lbGlucose} .\frac{1molGlucose}{180.156gGlucose} .\frac{6molCO_{2}}{1molGlucose} .\frac{44.01gCO_{2}}{1molCO_{2}} =305gCO_{2}

<em>How much light energy does it take to grow the 1.70 lb. sweet potato if the efficiency of photosynthesis is 0.86%?</em>

<em />

According to the enthalpy of the reaction, 2802.8 kJ are required to produce 1 mole of glucose. Then, for a 1.70 lb potato:

1.70lbPotato.\frac{27lbGlucose}{100lbPotato} .\frac{453.59gGlucose}{1lbGlucose} .\frac{1molGlucose}{180.156gGlucose} .\frac{2802.8kJ}{1molGlucose} .\frac{1}{0.86\% } =3.77\times 10^{5} kJ

7 0
3 years ago
Differences between selenium tellurium and polonium
dsp73

Explanation:

Selenium and tellurium are rare elements, respectively, in order of crustal abundance. However, polonium, on account of its radioactive decay, is less abundant. The main source of selenium and tellurium is the anode slime deposited during the electrolytic refining of copper.

5 0
3 years ago
Suppose you are titrating vinegar, which is an acetic acid solution of unknown strength, with a sodium hydroxide solution accord
OleMash [197]

Answer:

0.0626 M

Explanation:

Equation of the reaction

CH3COOH(aq) + NaOH(aq) ---------> CH3COONa(aq) + H2O

Concentration of acid CA= ???

Concentration of base CB= 0.1943 M

Volume of acid VA= 10.0ml

Volume of base VB= 32.22 ml

Number of moles of acid NA= 1

Number of moles of base NB= 1

From

CA= CB VB NA/VA NB

Hence ;

CA= 0.1943 M × 32.22 ml × 1/10.0ml ×1

CA= 0.0626 M

3 0
4 years ago
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