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Luba_88 [7]
3 years ago
12

a chemist reacts 150.0 grams of HCl with an excess of MnO2. if the following reaction occurs: MnO2+4HCl=MnCl2+2H2O+Cl2 how many

grams of mncl2 are formed
Chemistry
2 answers:
krok68 [10]3 years ago
4 0

Answer:

There are formed 98.05 g of MnCl₂

Explanation:

The reaction is this one:

MnO₂ + 4 HCl   →  MnCl₂ +  2 H₂O + Cl₂

First of all, determinate moles. Divide mass /molar mass

150 g / 36.45 g/m = 4.11 moles of HCl

Ratio between HCl and MnCl₂ is 4:1

4 moles of HCl produce 1 mol of Chloride

4.11 moles of HCl  'll produce (4.11 . 1)/ 4 =1.03 moles of chloride

Molar mass . Moles = Mass

Molar Mass MnCl₂ = 95.2 g/m

95.2 g/m  . 1.03 moles = 98.05 grams

Firdavs [7]3 years ago
3 0

Answer:

There are 129.4 grams of MnCl2 formed

Explanation:

Step 1: Data given

Mass of HCl = 150.0 grams

MnO2 = excess

Molar mass of HCl = 36.46 g/mol

Step 2: The balanced equation

MnO2+4HCl → MnCl2+2H2O+Cl2

Step 3: Calculate moles of HCl

Moles HCl = Mass HCl / molar mass HCl

Moles HCl = 150.0 grams / 36.46 g/mol

Moles HCl = 4.114 moles

Step 4: Calculate Moles of MnCl2

For 1 mol of MnO2 we need 4 moles of HCl to produce 1 mol of MnCl2, 2 moles of H2O and 1 mol Cl2

For 4.114 moles of HCl we'll have 4.114/4 = 1.0285 moles of MnCl2

Step 5: Calculate mass of MnCl2

Mass MnCl2 =moles MnCl2 * molar mass MnCl2

Mass MnCl2 = 1.0285 * 125.84 g/mol

Mass MnCl2 = 129.4 grams

There are 129.4 grams of MnCl2 formed

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Answer:

55.3 × 10²³ molecules

Explanation:

Given data:

Number of moles of C₁₁H₁₂O₂₂ = 9.18 mol

Number of molecules = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

For given data:

9.18 mol × 6.022 × 10²³ molecules /1 mol

55.3 × 10²³ molecules

5 0
3 years ago
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Assume that Na+ is being transported across a membrane via facilitated diffusion. Which of the following conditions would allow
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Answer:

<h2>4. Na+ diffusing toward the side of the membrane with Cl− and 50% less Na+.</h2>

Explanation:

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(Link; https://www.chemteam.info/SigFigs/SigFigsFable.html)
Anna007 [38]

1. The measurement turned out to be so expensive since the student did not rely on the significant figures to calculate the edge of the cube and approximate it to the next value (2.1 cm possibly) that would allow a simpler construction and therefore its cost was much lower .

2. The student had to take all the significant figures in his calculations, with which he would have:

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Volume = \frac{80g}{8.67g/mL}

Volume = 9.2272203 mL

Since the figure to be constructed is a cube, he had to calculate the cube root of the volume to find the value of the edge of the cube:

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Because the cube edge value is so specific, in order to manage his budget, he was able to order a 2.1 cm cube, which would bring the mass up to 80.29287 g, and in the lab reduce one of the faces to the appropriate weight. .

On the other hand, the main thing he had to do was ask how much it would cost to make a cube with those specifications, especially when they mentioned that it would be "expensive" and he only had $50.

The significant figures guarantee the correct operation of a machinery, a gear, a team in general, for which the accuracy will not only be taken to the millimeter, but sometimes microns or much more specific, as in the case of computer components, Therefore, it is very important, if not, and if arbitrary measurements are taken that do not consider significant figures, the components could not function properly, which would cause a loss in time, effort and manpower.

If you want to learn more about exercises with significant figures, you can see the next link: brainly.com/question/11904364?referrer=searchResults

4 0
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