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Shkiper50 [21]
3 years ago
11

Help I need a concluding sentence for my conclusion this is what i have so far: in conclusion we discovered that lamp oil has th

e most surface tension and polarity. We could fit the most drops of lamp oil on the penny without the bubble breaking. 34 drops was the maximum amount of drops we got. Lamp oil is a very polar substance
Chemistry
1 answer:
Liono4ka [1.6K]3 years ago
3 0
I personally, would remove " Lamp oil is a very polar substance " And replace it with what the article you wrote, was about. For example, lets say you wrote about why people love ice cream, you could have the ending sentence say. " And that's why I wrote about why people love ice cream" Or something along those lines. I'd write a few more facts about it first,  don't just have one sentence saying a fact without any supporting detail, hope this helps! :)
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What is the total number of atoms in 0.20 mol of propanone, CH3COCH3?
Nutka1998 [239]

Answer:

1.2×10²³ atoms.

Explanation:

Data obtained from the question include:

Mole of propanone = 0.20 mole

Number of atoms of propanone =.?

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ atoms.

This implies that 1 mole of propanone also contains 6.022×10²³ atoms.

Thus, we can obtain the number of atoms in 0.20 mole of propanone as illustrated below:

1 mole of propanone contains 6.022×10²³ atoms.

Therefore, 0.20 mole of propanone will contain = 0.2 × 6.022×10²³ = 1.2×10²³ atoms.

Thus, 0.20 mole of propanone contain

1.2×10²³ atoms.

6 0
3 years ago
Balance the following Chemical Equation:<br> NaBr +CaCl2-&gt; NaCl+ CaBr2
frozen [14]
First write all of the compounds/atoms in either side then fill in existing values and balance


Na- 1
Br- 1
Ca- 1
Cl- 2

Na- 1
Cl- 1
Ca-1
Br-2

Balance to get

2NaBr+CaCl2=2NaCl+CaBr2
7 0
3 years ago
If the half-life of a sample of a radioactive substance is 30 seconds, how much would be left after 60 seconds? A. one-fourth B.
Basile [38]
m=m_{0}*(\frac{1}{2})^{\frac{60}{30}}\\\\&#10;m=m_{0}*(\frac{1}{2})^{2}\\\\&#10;m=\frac{1}{4}m_{0}

If the half-life of a sample of a radioactive substance is 30 seconds, how much would be left after 60 seconds? <span>A. one-fourth</span>
7 0
3 years ago
Read 2 more answers
A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal
goblinko [34]

Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so <u>the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂</u>:

1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC

All of the hydrogens atoms in the compound ended up becoming H₂O, so <u>the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O</u>:

0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH

Because the compound is composed only by C, H and O, <u>the mass of O in the compound can be calculated by substraction</u>:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

  • mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C
  • mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H
  • mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

  • The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

8 0
3 years ago
Find the mass of 175.4mL of benzene if the density is 0.8786g/ml
bulgar [2K]
Hey there!:

Volume = 175.4 mL

density = 0.8786 g/mL  

mass = ?

Therefore:

D = m / V

0.8786 = m / 175.4

m = 0.8786 * 175.4

m = 154.10644 g

hope this helps!


5 0
3 years ago
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