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gavmur [86]
2 years ago
6

Explain the following observations

Chemistry
1 answer:
ddd [48]2 years ago
7 0

Answer:

arattarabajajanz h

Explanation:

taraaratqwtwusbnz

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Determine the number of grams of sodium carbonate needed to prepare 100.0 ml of a 2.5 m solution
SCORPION-xisa [38]

Answer:

26.5 g

Explanation:

First we convert 100.0 mL to L:

  • 100.0 mL / 1000 = 0.100 L

Now we <u>calculate how many moles of sodium carbonate are needed</u>, using the <em>definition of molarity</em>:

  • Molarity = moles / liters
  • moles = molarity * liters
  • 2.5 M * 0.100 L = 0.25 mol

Finally we <u>convert 0.25 moles of sodium carbonate into grams</u>, using its <em>molar mass</em>:

  • 0.25 mol * 106 g/mol = 26.5 g
6 0
2 years ago
Asap... define pure substances...​
gtnhenbr [62]

A pure substance refers to an element or a compound that has no component of another compound or element. Pure substances are made of only one type of atom or molecule. Hydrogen gas and pure iron are examples of pure substances. Hydrogen consists of hydrogen atoms only while iron consists of only iron atoms. Mixing two pure substances results in a mixture. To separate the two, scientists use a method known as filtration. Mixtures can either be homogeneous or heterogeneous. The measure used to determine how pure a substance may be called purity. Besides hydrogen and iron, other pure substances include gold, diamonds, sugar, and baking soda.

3 0
3 years ago
Read 2 more answers
What are the steps to evaporation but without heat just evaporation steps to follow
densk [106]
First, the sun shines liquid (ocean) Next, the water evaporates 
8 0
3 years ago
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What mass of precipitate forms when 185.5 ml of 0.533 m naoh is added to 627 ml of a solution that contains 15.8 g of aluminum s
Law Incorporation [45]

Answer:

2,57 g of precipitate.

Explanation:

For the reaction:

6 NaOH + Al₂(SO₄)₃ → 2 Al(OH)₃ + 3 Na₂SO₄

The precipitate is Al(OH)₃.

185,5mL of 0,533M NaOH are:

0,1855L × 0,533M = <em>0,0989 moles NaOH</em>

Moles of Al₂(SO₄)₃ are:

15,8g × \frac{1mol}{342,15g} = <em>0,0462 moles Al₂(SO₄)₃</em>

For the total reaction of 0,0989 moles NaOH with Al₂(SO₄)₃ you need:

0,0989moles NaOH × \frac{1molAl_{2}(SO_{4})_{3}}{6 moles NaOH} = <em>0,0165 moles Al₂(SO₄)₃</em>

As you have <em>0,0462 moles Al₂(SO₄)₃ </em>the limiting reactant is NaOH.

0,0989 moles of NaOH produce:

0,0989moles NaOH × \frac{2molAl(OH)_{3}}{6 moles NaOH} = <em>0,0330 moles of Al(OH)₃</em>

These moles are:

0,0330 moles of Al(OH)₃ × (78 g/mol) = <em>2,57 g of Al(OH)₃ ≡ mass of precipitate</em>

<em></em>

I hope it helps!

<em> </em>

3 0
3 years ago
And a 40N Torce in a southeriy direction?
dimaraw [331]

Answer:

I Don't Know.

Explanation: I'm Sorry, But I'm Not Good At This Subject. Atleast I'm Being Honest.

8 0
3 years ago
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