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choli [55]
3 years ago
14

What is not consider a property of substance

Chemistry
1 answer:
nasty-shy [4]3 years ago
4 0

Answer:

f

Explanation:

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If there is a graduated cylinder containing 20mL of water and a small rock is gently placed inside it, and the water level insid
LekaFEV [45]

Answer:

5

Explanation:

just subtract 20-25 and 5 is left over which is how much the water when up with the rock.

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What is chemistry?????​
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Answer:

chemistry - the science that studies the properties of substances and natural fenomens .

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C) A football feels hard on a hot day but feels softer in the evening​
Mariana [72]

Answer: ^_^

Explanation:

its because during day time the ball becomes heated up and the air inside it try to come out as the heat air always try to go up swelling up the ball,

so in the evening when the air cools down the ball too cools down and air inside it also cools down making the ball feel soft,

Therefore, a football feels hard on a hot day but feels softer in the evening​

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3 years ago
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Coal containing 15.0% H2O, 2.0% S and 83.0% C by mass is burnt with the stoichiometric amount of air in a furnace. What is the m
devlian [24]

Answer:

This is a coal combustion process and we will assume

Inlet coal amount = 100kg

It means that there are

15kg of H2O, 2kg of Sulphur and 83kg of Carbon

Now to find the mole fraction of SO2(g) in the exhaust?

Molar mass of S = 32kg/kmol

Initial moles n  of S = 2/32 = 0.0625kmols

Reaction:  S + O₂ = SO₂

That is 1 mole of S reacts with 1 mole of O₂ to give 1 mole of SO₂

Then, it means for 0.0625 kmoles of S, we will have 0.0625 kmole of SO2 coming out of the exhaust

The mole fraction of SO2(g) in the exhaust=0.0625kmols

Explanation:

5 0
3 years ago
How many moles are in 3.113 g of Au?Molar mass of Au=197 g/mol
SVEN [57.7K]
<h3>Answer:</h3>

0.0157 g Au

<h3>General Formulas and Concepts:</h3>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.113 g Au

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

Molar Mass of Au - 197.87 g/mol

<u>Step 3: Convert</u>

<u />3.113 \ g \ Au(\frac{1 \ mol \ Au}{197.87 \ g \ Au} ) = 0.015733 g Au

<u>Step 4: Check</u>

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

0.015733 g Au ≈ 0.0157 g Au

3 0
3 years ago
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