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Musya8 [376]
3 years ago
11

Identify two kinds of data the student will likely collect during his experiment. Identify the type of data collected and how th

e student might present his data.
Chemistry
2 answers:
-BARSIC- [3]3 years ago
6 0
Qualitative (physical e. g. colour) and quantitative (numerical e. g. 3 grams) data
DiKsa [7]3 years ago
4 0

The two kinds of data the student is most likely to collect are qualitative and quantitative data. This means that he will gather data based on his senses (qualitative). This could refer to things like the size of the tomatoes. The student also could gather numerical data (Quantitative). This is referring to how many tomatoes were bigger because of the increase sunlight exposure. This student could present this information in pieces like graphs and charts to show the quantitative, and models and pictures to how the qualitative data.

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Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 3.43 g of octane is m
Lisa [10]

Answer:

The maximum mass of carbon dioxide, CO₂ produced is 2.64 g

Explanation:

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

2C₈H₁₈ + 25O₂ —> 16CO₂ + 18H₂O

Next, we shall determine the masses of C₈H₁₈ and O₂ that reacted and the mass of CO₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of C₈H₁₈ = (8×12) + (18×1)

= 96 + 18

= 114 g/mol

Mass of C₈H₁₈ from the balanced equation = 2 × 114 = 228 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 25 × 32 = 800 g

Molar mass of CO₂ = 12 + (2×16)

= 12 + 32

= 44 g/mol

Mass of CO₂ from the balanced equation = 16 × 44 = 704 g

SUMMARY:

From the balanced equation above,

228 g of C₈H₁₈ reacted with 800 g of O₂ to produce 704 g of CO₂.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

228 g of C₈H₁₈ reacted with 800 g of O₂.

Therefore, 3.43 g of C₈H₁₈ will react with = (3.43 × 800)/228 = 12.04 of O₂

From the calculations made above, we can see that a higher mass (i.e 12.04 g) of O₂ than what was given (i.e 3 g) is needed to react completely with 3.43 g of C₈H₁₈. Therefore, O₂ is the limiting reactant and C₈H₁₈ is the excess reactant.

Finally, we shall determine the maximum mass of carbon dioxide, CO₂ produced from the reaction.

To obtain the maximum mass of carbon dioxide, CO₂ produced, the limiting reactant will be used because all of it is consumed in the reaction.

The limiting reactant is O₂ and the maximum mass of carbon dioxide, CO₂ produced can be obtained as follow:

From the balanced equation above,

800 g of O₂ reacted to produce 704 g of CO₂.

Therefore, 3 g of O₂ will react to produce = (3 × 704)/800 = 2.64 g of CO₂.

Thus, the maximum mass of carbon dioxide, CO₂ produced is 2.64 g

8 0
3 years ago
A sample of CO2 gas at 100 degrees Celsius has a volume of 250 mL at 760 mm Hg. How many moles of CO2 are present
Fed [463]

There are 8.16 × 10-³ moles of CO2 gas at 100°C with a volume of 250 mL at 760 mm Hg.

HOW TO CALCULATE NUMBER OF MOLES:

The number of moles of a sample of gas can be calculated using the following formula:

PV = nRT

Where;

  • P = pressure of gas (atm)
  • V = volume (L)
  • n = number of moles (mol)
  • R = gas law constant (0.0821 Latm/molK)
  • T = temperature (K)

According to this question;

  • P = 760mmHg = 1 atm
  • T = 100°C = 100 + 273 = 373K
  • V = 250mL = 0.250L
  • n = ?

1 × 0.250 = n × 0.0821 × 373

0.250 = 30.62n

n = 0.250 ÷ 30.62

n = 8.16 × 10-³mol

Therefore, there are 8.16 × 10-³ moles of CO2 gas at 100°C with a volume of 250 mL at 760 mm Hg.

Learn more about number of moles at: brainly.com/question/4147359

6 0
3 years ago
what organelle is found only in plant cells, provides structure and support outside of cell, and protect the plant and provide r
fgiga [73]

Answer:

Cell Wall

Explanation:

5 0
4 years ago
Read 2 more answers
Express the concentration of a 0.0390 M aqueous solution of fluoride, F–, in mass percentage and in parts per million. Assume th
mart [117]
<span>ppm = mg/L
or
Kg solution Mass % = Mass of Solute/Mass of solution

0.0390 M = x moles/1L
x = 0.0390 Moles of F-
0.0390 mol x 18.998403 g/mol = 0.7409 g

PPM:
1mg = 1000g
0.7409g = 740.9mg
740.9mg/1L = 740.9mg

Mass Percent:
density 1 g/mL
thus
1 g = 1 mL
thus 1 L = 1 kg
thus 1 L = 1000 g
so,
(0.7409 g/1000 g)*100
 = 0.07409% (m/m)</span>
4 0
3 years ago
What is the purpose of writing Shorthand Electron Configuration Using Noble Gases?<br> 30 POINTS!
aivan3 [116]

The noble gas configuration system helps to shorten the total electron configuration by using the symbol for the noble gas that lies before the particular element in the periodic table.  

For example:

Na has atomic number 11

Electron configuration of Na: 1s2 2s2 2p6 3s1

Noble gas electron configuration of Na: [Ne] 3s1

Ne has atomic number 10 and lies before Na in the periodic table.  


6 0
3 years ago
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