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zysi [14]
2 years ago
11

How many atoms are contained in 3.46 moles of magnesium

Chemistry
1 answer:
Liono4ka [1.6K]2 years ago
6 0

Answer:

2.08 * 10^{24}

Explanation:

Avagadros number times the # of moles.

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Some commercially available algaecides for swimming pools claim to contain 7% copper. Could the method used in this experiment t
AleksandrR [38]

Answer:

Explanation:

  1. 7% copper implies 7 w/v%(weight/volume %) of copper. This implies a 100 mL algaecide arrangement contains 7 grams of copper.  
  2. Henceforth we have a thought of the measure of copper that ought to be available in a given example of algaecide.  
  3. Indeed, even a 1 mL algaecide test is sufficient to discover the percentage of copper in it.  
  4. so a 25mL sample of algaecide must have around 1.75g of copper.
4 0
3 years ago
The article is on newsela called the water cycle
garri49 [273]

Answer:

Explanation:Canguru comeu algumas folhas de 3 galhos de eucalipto. Cada galho tinha inicialmente20 folhas. Canguru comeu algumas folhas do primeiro galho e depois comeu tantas folhas do segundo galho quantas tinham sido deixadas no primeiro galho. Depois ele comeu 2 folhas do terceiro galho. No total, quantas folhas foram dei-xadas nos 3 galhos?Canguru comeu algumas folhas de 3 galhos de eucalipto. Cada galho tinha inicialmente20 folhas. Canguru comeu algumas folhas do primeiro galho e depois comeu tantas folhas do segundo galho quantas tinham sido deixadas no primeiro galho. Depois ele comeu 2 folhas do terceiro galho. No total, quantas folhas foram dei-xadas nos 3 galhos?Canguru comeu algumas folhas de 3 galhos de eucalipto. Cada galho tinha inicialmente20 folhas. Canguru comeu algumas folhas do primeiro galho e depois comeu tantas folhas do segundo galho quantas tinham sido deixadas no primeiro galho. Depois ele comeu 2 folhas do terceiro galho. No total, quantas folhas foram dei-xadas nos 3 galhos?Canguru comeu algumas folhas de 3 galhos de eucalipto. Cada galho tinha inicialmente20 folhas. Canguru comeu algumas folhas do primeiro galho e depois comeu tantas folhas do segundo galho quantas tinham sido deixadas no primeiro galho. Depois ele comeu 2 folhas do terceiro galho. No total, quantas folhas foram dei-xadas nos 3 galhos?Canguru comeu algumas folhas de 3 galhos de eucalipto. Cada galho tinha inicialmente20 folhas. Canguru comeu algumas folhas do primeiro galho e depois comeu tantas folhas do segundo galho quantas tinham sido deixadas no primeiro galho. Depois ele comeu 2 folhas do terceiro galho. No total, quantas folhas foram dei-xadas nos 3 galhos?

7 0
3 years ago
What is the balanced form of the following equation? Br 2 + S 2 O 3 2– + H 2 O → Br 1– + SO 4 2– + H +
nikitadnepr [17]

Answer:

4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

Explanation:

Br₂ +  S₂O₃²⁻  + H₂O  → Br⁻ + SO₄²⁻ + H⁺

This is a redox reaction:

Br₂ changes the oxidation state from 0 to -1, so it was reduced

In the S₂O₃⁻² anion S changes the oxidation state from +2 to +6 in sulfate anion. (S₂O₃⁻², it is called thiosulfate)

We have protons in the main equation, so we assume we are in acidic medium:

Br₂ + 2e⁻ → 2Br⁻         Reduction

We balanced the bromide with 2, so the bromine has gained 2 electrons.

<u>5H₂O</u> + S₂O₃²⁻ → 2SO₄²⁻ + <u>10H⁺</u> + <em>8e</em>-  Oxidation

First of all, we add 2 to the sulfate anion in the product side, in order to balance the S.

As we have 8 O in right side, and 3 O in left side, we must add 5 O. We add 5 water in the place where the O are lower (reactant side).

Now, we have 10 H, in the reactant side, so we balance the product side with protons (10 H⁺).

Sulfur changed the oxidation state from +2 to +6, so it released 4 electrons, but, if you see thiosulfate anion you have 2 sulfurs so finally it has released 8 electrons.

Electrons are unbalanced so we multiply reduction x4, and oxidation x1.

(Br₂ + 2e⁻ → 2Br⁻) . 4 = 4Br₂ + 8e⁻ → 8Br⁻

(5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + <em>8e</em>-) . 1 = STAYS THE SAME.

We sum both half reactions, to cancel the elecetrons:

4Br₂ + 8e⁻ + 5H₂O + S₂O₃²⁻  → 2SO₄²⁻ + 10H⁺ + <em>8e</em>- + 8Br⁻

Finally the balanced reaction is: 4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

5 0
3 years ago
NaOH  + H2SO4 =
mariarad [96]
2NaOH+H_2SO_4\Rightarrow\ Na_2SO_4+2H_2O\\&#10;NaOH+HNO_3\Rightarrow\ NaNO_3+H_2O\\&#10;Cu(OH)_2+2HCl\Rightarrow\ CuCl_2+2H_2O\\&#10;Cu(OH)2+H_2SO_4\Rightarrow\ CuSO_4+2H_2O\\&#10;Cu(OH)_2+2HNO_3\Rightarrow\ Cu(NO_3)_2+2H_2O
3 0
2 years ago
How many grams of nitric acid will react completely with a block of iron metal that is 4.5 cm by 3.0 cm by 3.5 cm, if the densit
denpristay [2]

Balance the equation first: 

2 Fe+6 HNO3→2 Fe(NO3)3+3H2

Then calculate mass of Iron :

4.5×3.0×3.5 cm3(1 mL1 cm3)(7.87 g Fe1 ml)=371.86 g Fe

Now use Stoichiometry:

371.86 g Fe×(1 mol Fe55.85 g Fe)×(6 mol HNO32 mol Fe)=19.97 mol HNO3

Convert moles of nitric acid to grams

19.97 mol HNO3×(63.01 g HNO31 mol HNO3)=1258.3 g HNO3



7 0
2 years ago
Read 2 more answers
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