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Sholpan [36]
2 years ago
15

On easter island what was the primary ecological disaster and why did the people not prevent it?

Chemistry
1 answer:
Amiraneli [1.4K]2 years ago
4 0

Answer:Easter Islanders Didn't Cause Ecological Disaster on Their Island, New Research Finds. New research provides evidence contrary to the widely-held belief that the prehistoric population of Easter Island (Rapa Nui) caused an ecological disaster through rapid deforestation and over-exploitation of natural resources.

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A solution is made by dissolving 9.74 g of sodium sulfate in water to a final volume of 165 mL of solution. What is the weight/w
MatroZZZ [7]

Answer: The weight/weight % or percent by mass of the solute is 5.41 %.

Explanation:

Mass of the sodium sulfate,w = 9.74 g

Volume of the water = 165 mL

Density of the water = 1 g/mL

Density=1 g/mL=\frac{\text{Mass of water}}{\text{Volume of water}}

Mass of the water =1 g/mL\times 165 mL=165 g

Mass of the solution, W:

Mass of solute + Mass of solvent =9.47 g + 165 g=174.47 g

w/w\%=\frac{w\times 100}{W}=\frac{9.45 g\times 100}{174.47 g}=5.41 \%

The weight/weight % or percent by mass of the solute is 5.41 %.


8 0
3 years ago
I need help solving this chemistry question ​
Arturiano [62]

Answer:

I think the answer is 22.2

Explanation: What i DID was adding 12.0 + 10.0 and than that gave me 22 the I had added the to 0.200 and that how i got 22.2. Sorry if i got is wrong. :(

3 0
3 years ago
A sample of sugar (C12H22O11) contains
Oduvanchick [21]

Answer: 0.25 mol

Explanation:

Use the formula n=N/NA

n= number of mols

N =  number of particles

Nᵃ = Avogadros constant = 6.02 x

So, n=

The 10 to the power of 23 cancels out and you are left with 1.505/6.02, which is approximately 1/4. This is the same as 0.25 mol.

Hope this helped :)

5 0
3 years ago
A sample of chromium oxide is 76.5hromium by weight. what is the simplest formula of the oxide?
otez555 [7]

CrO and Cr₂O₃ make up the simplest chromium oxide formula.

What name does Cr₂O₃ use?

  • Chromium oxide (Cr₂O₃)sometimes referred to as chromium sesquioxide or chromic oxide, is a compound in which chromium is oxidized to a +3 state. Sodium dichromate is calcined with either carbon or sulfur to produce it.
  • Eskolaite, a mineral that bears the name of the Finnish geologist Pentti Eskola, is a kind of chromium oxide green that may be found in nature. The metallic glassy green surface of this unusual material has an unsettling moss-like look that may be used to conceal oneself in the environment.
  • Studies on humans have conclusively shown that chromium (VI) breathed is a potential carcinogen, increasing the likelihood of developing lung cancer. According to animal studies, chromium (VI) exposure by inhalation can result in lung cancers.

Learn more about chromium here:

brainly.com/question/15588080

#SPJ4

6 0
1 year ago
18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c
Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

   Percentage by mass of hydrogen: 0.77%

b)  Percentage by mass of chlorine: 80.37%

c) Molecular formula: C_{2} H Cl_{3}

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}

The same process is used to calculate the amount of hydrogen:

0.089g H_{2}O*\frac{2g H}{18g H_{2}O }  = 0.010g H

The percentage by mass of carbon and hydrogen are calculated as follows:

%C\frac{0.238g}{1.3g} *100%= 18.3%

%H\frac{0.010g}{1.3g} *100%=0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}= 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl=\frac{0.43g}{0.535g} * 100%= 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

18.3g C*\frac{1mol C}{12g C} = 1.52mol C

0.77g H*\frac{1mol H}{1g H} = 0.77mol H

80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl

Dividing each of the quantities above by the smallest (0.77mol), the  subscripts in a tentative formula would be

C=\frac{1.52}{0.77} = 1.97 ≈ 2

H = \frac{0.77}{0.77} = 1

Cl =\frac{2.27}{0.77}=2.94≈3

The empirical formula for the compound is:

C_{2} H Cl_{3}

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0
3 years ago
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