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nikklg [1K]
2 years ago
14

3. КОН + — — Н3РО4 1 + K3PO4 — Н,0 —

Chemistry
1 answer:
expeople1 [14]2 years ago
3 0
What does this mean?
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Chalcopyrite is an ore with the composition cufes2. what is the percentage of iron in a 39.6 g sample of this ore? answer in uni
iren2701 [21]
<span>Well it depends on percentage by what, but I'll just assume that it's percentage by mass. For this, we look at the atomic masses of the elements present in the compound. Cu has an atomic mass of 63.546 amu Fe has 55.845 amu and S has 36.065 amu Since there are 2 molecules of Sulfur for each one of Cu and Fe, we'll multiply the Sulfur atomic weight by 2 to obtain 72.13 amu So we have not established the mass of the compound in amus 63.546 + 55.845 + 72.13 = 191.521 That is the atomic mass of Chalcopyrite. and Iron's atomic mass is 55.845 So to get the percentage, or fraction of iron, we take 55.845 / 191.521 Which comes out to 29.15% by mass Mass of the sample is not needed for this calculation, but since the question mentions it I would go ahead and check if the question isn't also asking for the mass of Iron in the sample as well, in which case you just find the 29.15% of 67.7g</span>
5 0
3 years ago
What would happen to earth's orbit around the sun if there were gravity
kipiarov [429]
If the sun was not there the earth would travel in a straight line
7 0
3 years ago
How many moles in 30.0 grams of h3po4
jok3333 [9.3K]
Molar mass H₃PO₄ = 98.0 g/mol

1 mole ----- 98.0 g
? mole ------ 30.0 g

moles = 30.0 * 1 / 98.0

= 0.306 moles

hope this helps!


5 0
3 years ago
What is the volume of 40.0 grams of argon gas at STP ?
MrRa [10]

Answer:

24.9 L Ar

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Aqueous Solutions</u>

  • States of Matter

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 40.0 g Ar

[Solve] L Ar

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of Ar - 39.95 g/mol

[STP] 22.4 L = 1 mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 40.0 \ g \ Ar(\frac{1 \ mol \ Ar}{39.95 \ g \ Ar})(\frac{22.4 \ L \ Ar}{1 \ mol \ Ar})
  2. [DA] Divide/Multiply [Cancel out units]:                                                         \displaystyle 24.9235 \ L \ Ar

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

24.9235 L Ar ≈ 24.9 L Ar

5 0
3 years ago
What explains the fact that no machine is 100 percent efficient?
nydimaria [60]
Because some machine might not work and would give you the wrong answer
4 0
3 years ago
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