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Reptile [31]
4 years ago
7

Which natural resource is limited in supply?

Chemistry
2 answers:
Dominik [7]4 years ago
8 0
It’s b; petroleum , since it’s an oil i’m pretty sure , which is limited
denpristay [2]4 years ago
3 0

Answer:

The correct answer is the lasf one solar energy

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CHEMISTRY:There are 9 rotons & 11 neutrons. What is the atomic number & atomic mass?
ziro4ka [17]

Hey there!

Atomic number is the number of protons.

Atomic mass is the sum of protons and neutrons.

9 + 11 = 20

Your answer is B) The atomic number is 9 & the atomic mass is 20.

Hope this helps!

5 0
3 years ago
What is 1 item that is a compound?<br><br>Give an example of how the 1 item is a compound :D
spayn [35]

Water

Water is a compound because it is made from more than one kind of element (oxygen and hydrogen).

3 0
4 years ago
Scientist A produces 83.67 g KMnO4 while Scientist B produces 81.35 g KMnO4.
noname [10]

Answer:

Y_A=92.1\%\\\\Y_B=89.6\%

Explanation:

Hello there!

In this case, according to the given chemical equation for the reaction for the production of potassium permanganate, we can see a 2:2 mole ratio of this product to the starting manganese (II) oxide, which means, we can calculate the theoretical yield of the former via stoichiometry:

m_{KMnO_4}=50.0gMnO_2*\frac{1molMnO_2}{86.94gMnO_2}*\frac{2molKMnO_4}{2molMnO_2}  *\frac{158.034gKMnO_4}{1molKMnO_4} \\\\m_{KMnO_4}=90.9gKMnO_4

Now, we are able to compute the percent yields, by using the actual yield each scientist got:

Y_A=\frac{83.67g}{90.9g} *100\%=92.1\%\\\\Y_B=\frac{81.35g}{90.9g} *100\%=89.6\%

Regards!

8 0
3 years ago
How many mols is 1433.6 iron atoms
larisa86 [58]
<span>How many mols is 1433.6 iron atoms? my calculations are 2.380548514178884521</span>

7 0
3 years ago
A 1.0857 gram pure sample of a compound containing only carbon, hydrogen, and oxygen was burned in excess oxygen gas. 2.190 g of
Goryan [66]

Answer:

  • C₂ H₄ O

Explanation:

<u>1) Mass of carbon (C) in 2.190 g of carbon dioxide (CO₂)</u>

  • atomic mass of C: 12.0107 g/mol
  • molar mass of CO₂: 44.01 g/mol
  • Set a proportion: 12.0107 g of C / 44.01 g of CO₂ = x / 2.190 g of CO₂
  • Solve for x:

         x = (12.0107 g of C / 44.01 g of CO₂ ) × 2.190 g of CO₂ = 0.59767 g of C

<u />

<u>2) Mass of hydrogen (H) in 0.930 g of water (H₂O)</u>

  • atomic mass of H: 1.00784 g/mol
  • molar mass of H₂O: 18.01528 g/mol
  • proportion: 2 × 1.00784 g of H / 18.01528 g of H₂O = x / 0.930 g of H₂O
  • Solve for x:

        x = ( 2 × 1.00784 g of H / 18.01528 g of H₂O) × 0.930 g of H₂O = 0.10406 g of H

<u>3) Mass of oxygen (O) in 1.0857 g of pure sample</u>

  • Mass of O = mass of pure sample - mass of C - mass of H
  • Mass of O = 1.0857 g - 0.59767 g - 0.10406 = 0.38397 g O

Round to four decimals: Mass of O = 0.3840 g

<u>4) Mole calculations</u>

Divide the mass in grams of each element by its atomic mass:

  • C: 0.59767 g / 12.0107 g/mol = 0.04976 mol
  • H: 0.10406 g / 1.00784 g/mol = 0.10325 mol
  • O: 0.3840 g / 15.999 g/mol = 0.02400 mol

<u>5) Divide every amount by the smallest value (to find the mole ratios)</u>

  • C: 0.04976 mol / 0.02400 mol = 2.07 ≈ 2
  • H: 0.10325 mol / 0.02400 mol = 4.3 ≈ 4
  • O: 0.02400 mol / 0.02400 mol = 1

Thus the mole ratio is 2 : 4 : 1, and the empirical formula is:

  • <u>C₂ H₄ O </u>← answer
3 0
3 years ago
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