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vladimir1956 [14]
3 years ago
5

Help me please i need the answer now(nonsense report)​

Chemistry
1 answer:
rusak2 [61]3 years ago
3 0

Answer:

Quantitative

  • Words, Measurement, Numbers, Charts, Instruments, Quantity, Data

Qualitative

  • Touch, Quality, Hear, Senses, Description, Exact, Narrative, Analyze

Explanation :

Quantitative Data

  • Can't measure
  • Has numerical value
  • Can campare
  • Do not depend on human eye

Qualitative Data

  • Can't measure
  • Don't have numerical value
  • Can't campare
  • Depend on human eye
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Calculate the energy required to ionize a hydrogen atom to an excited state where the electron is initially in the n = 5 energy
SVETLANKA909090 [29]

. The energy of shells in a hydrogen atom is calculated by the formula E = -Eo/n^2 where n is any integer, and Eo = 2.179X10^-18 J. So, the energy of a ground state electron in hydrogen is:

E = -2.179X10^-18 J / 1^2 = -2.179X10^-21 kJ

Consequently, to ionize this electron would require the input of 2.179X10^-21 kJ


2. The wavelength of a photon with this energy would be:

Energy = hc/wavelength

wavelength = hc/energy

wavelength = 6.626X10^-34 Js (2.998X10^8 m/s) / 2.179X10^-18 J = 9.116X10^-8 m

Converting to nanometers gives: 91.16 nm


3. Repeat the calculation in 1, but using n=5.


4. Repeat the calculation in 2 using the energy calculated in 3.

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3 years ago
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Why does the moon spin around the earth and the earth spins around the sun
liberstina [14]

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3 years ago
Two samples of the same compound are compared. what does the data represent? sample 1: 24.22 g carbon and 32.00 g oxygen sample
goblinko [34]

According to law of definite proportion:

In a compound, elements are always arranged in fixed ratio by mass.

Here, sample 1 has 23.22 g Carbon and 32.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

n_{C}=\frac{m_{C}}{M_{C}}=\frac{24.22 g}{12 g/mol}\approx 2 mol

Similarly, number of moles of oxygen will be:

n_{O}=\frac{m_{O}}{M_{O}}=\frac{32 g}{16 g/mol}=2 mol

The ratio of number of moles of carbon and oxygen will be:

C:O=n_{C}:n_{O}=2:2=1:1

Therefore, formula of compound will be CO.

Sample 2:

It has 36.22 g Carbon and 48.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

n_{C}=\frac{m_{C}}{M_{C}}=\frac{36.22 g}{12 g/mol}\approx 3 mol

Similarly, number of moles of oxygen will be:

n_{O}=\frac{m_{O}}{M_{O}}=\frac{48 g}{16 g/mol}=3 mol

The ratio of number of moles of carbon and oxygen will be:

C:O=n_{C}:n_{O}=3:3=1:1

The formula of compound will be CO.

Therefore, it is proved that carbon and oxygen are present in fixed ratios in both the samples.


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3 years ago
Enter the appropriate symbol for an isotope of potassium-39 corresponding to the isotope notation A/Z x
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The balanced equation for combustion in an acetylene torch is shown below: 2C2H2 + 5O2 → 4CO2 + 2H2O The acetylene tank contains
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Answer: 67.2 moles

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According to the given balanced equation, 2 moles of acetylene C_2H_2 combine with 5 moles of oxygen O_2  to produce 4 moles of carbon dioxide CO_2.

Thus if 2 moles of acetylene C_2H_2 combine with = 5 moles of oxygen O_2

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But as only 84 moles of oxygen are available, acetylene is not a limiting reagent.

5 moles of oxygen O_2 reacts with = 2 moles of acetylene C_2H_2

84 moles of oxygen O_2 reacts with=\frac{2}{5}\times {84}=33.6 moles of acetylene C_2H_2

Thus Oxygen is the limiting reagent as it limits the formation of products. Acetylene is excess reagent as it is present in excess.

2 moles of acetylene C_2H_2  produce= 4 moles of carbon dioxide CO_2.

33.6 moles of acetylene C_2H_2 produce=\frac{4}{2}\times {33.6}=67.2 moles of of carbon dioxide CO_2.


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3 years ago
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