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Mkey [24]
3 years ago
13

1. Magnesium chloride solution reacts with silver nitrate solution to form magnesium nitrate

Chemistry
1 answer:
Whitepunk [10]3 years ago
6 0

a. 1,4332 g

b. 7.54~g

<h3>Further explanation</h3>

Given

Reaction

MgCl2 (s) + 2 AgNO3 (aq) → Mg(NO3)2 (aq) + 2 AgCl (s)

20 cm of 2.5 mol/dm^3 of MgCl2

20 cm of 2.5 g/dm^3 of MgCl2

Required

the mass of silver chloride - AgCl

Solution

a. mol MgCl2 :

\tt 20~cm^3=20\times 10^{-3}~dm^3\\\\mol=M\times V\\\\mol=2.5~mol/dm^3\times 20\times 10^{-3}DM^3=0.05

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.05=0.1

mass AgCl(MW=143,32 g/mol)= 0.1 x 143,32=1,4332 g

b. mol MgCl2 (MW=95.211 /mol):

\tt mol=M\times V\\\\mol=\dfrac{2.5~g/dm^3}{95,211 g/mol}=0.0263~mol/dm^3

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.0263=0.0526

mass AgCl(MW=143,32 g/mol)= 0.0526 x 143,32=7.54~g

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When an electron in a 2p orbital of a particular atom makes a transition to the 2s orbital, a photon of approximate wavelength 6
Mariulka [41]

Answer:

The energy difference between these 2p and 2s orbitals is 3.07\times 10^{-19} J

Explanation:

Wavelength of the photon emitted = \lambda =646.3 nm =646.3\times 10^{-9} m

Energy of the photon will corresponds to the energy difference between 2p and 2s orbital = E

Energy of the photon is given by Planck's equation:

E=\frac{hc}{\lambda }

h = Planck's constant = 6.626\tiomes 10^{-34} Js

c = Speed of the light = 3\times 10^8 m/s

E=\frac{6.626\tiomes 10^{-34} Js\times 3\times 10^8 m/s}{646.3\times 10^{-9} m}

E=3.07\times 10^{-19} J

The energy difference between these 2p and 2s orbitals is 3.07\times 10^{-19} J

3 0
3 years ago
Why did mendeleev leave blank spots in his periodic table?
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He thought elements that haven't been discovered belonged in the place of the gap. He could also use the atomic mass of the missing elements
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3 years ago
2.what is the purpose of the sodium sulfate? (1 pt) why did you rinse the sodium sulfate with an additional portion of methylene
suter [353]

To ensure that the 9-Fluorenone was totally dry, it had to be washed with methylene chloride. To make sure that methylene chloride is present in a pure solution, sodium sulfate binds to water and precipitates.

<h3>What is the purpose of the sodium sulfate?</h3>
  • Although it has numerous additional uses, sodium sulfate is primarily employed in the production of detergents and in the Kraft process of paper pulping.
  • The decahydrate's natural mineral form, mirabilite, accounts for about half of the world's output, with the other half coming from chemical byproducts. Sodium sulfate was used as a drying, isolating, and anhydrous salt for the 9-fluorenone.
  • To make sure that methylene chloride is present in a pure solution, sodium sulfate binds to water and precipitates.
  • The sodium salt of sulfuric acid is known as sodium sulfate. Na2SO4 is the chemical formula for sodium sulfate. The mineral thenardite, which is also known as anhydrous sulfate, is described as a white, crystalline solid, whereas the decahydrate Na2SO4. 10H2O is also known as Glauber's salt or the mirabilis salt.

To know more about sodium sulfate, refer:

brainly.com/question/23509646

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4 0
1 year ago
How many miles can you drive, if your car gets 21.3 miles per gallon and you have 12.0 gallons of gas?
Travka [436]

Answer:

255.6

Explanation:

If you have 12 gallons and get 21.3mpg,

-Multiply 21.3 by 12

-you can travel 255.6 miles before running out of gas.

-If you need to estimate, round up to 256 miles.

6 0
3 years ago
The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 Fe2(SO4)3 + 2Na3PO4 What is the t
slava [35]

<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of iron(III) phosphate = 20.00 g

Molar mass of iron(III) phosphate = 150.82 g/mol

Putting values in equation 1, we get:

\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol

The given chemical equation follows:

2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4

As, sodium sulfate is present in excess. So, it is considered as an excess reagent.

Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate

So, 0.133 moles of iron(III) phosphate will produce = \frac{1}{2}\times 0.133=0.0665moles of iron(III) sulfate

Now, calculating the mass of iron(III) sulfate from equation 1, we get:

Molar mass of iron(III) sulfate = 399.9 g/mol

Moles of iron(III) sulfate = 0.0665 moles

Putting values in equation 1, we get:

0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g

Hence, the theoretical yield of iron(III) sulfate is 26.6 grams

8 0
3 years ago
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