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SSSSS [86.1K]
3 years ago
5

For the reaction SO3 + H2O + H2SO4, how many grams of sulfuric acid can be

Chemistry
1 answer:
dolphi86 [110]3 years ago
8 0

Answer:

Mass = 544 g

Explanation:

Given data:

Mass of sulfuric acid formed = ?

Mass of water react = 100.0 g

Solution:

Chemical equation:

SO₃ + H₂O     →     H₂SO₄

Number of moles of water:

Number of moles = mass/molar mass

Number of moles = 100 g/ 18 g/mol

Number of moles = 5.5 mol

Now we will compare the moles of water and H₂SO₄.

                  H₂O       :         H₂SO₄

                     1          :            1

                 5.56       :           5.56

Mass of H₂SO₄:

Mass = number of moles × molar mass

Mass = 5.56 mol × 98.0 g/mol

Mass = 544 g

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How many moles of calcium chloride (CaCl2) are needed to react completely with 6.2 moles of silver nitrate (AgNO3)?
soldier1979 [14.2K]

Answer:

The no. of moles of CaCl₂ are needed = 3.1 mol.

Explanation:

  • From the balanced reaction:

<em>2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃)₂,</em>

It is clear that 2 mol of AgNO₃ react with 1 mol of CaCl₂ to produce 2 mol of AgCl and 1 mol of Ca(NO₃)₂.

<u><em>Using cross multiplication:</em></u>

2 mol of AgNO₃ need  → 1 mol of CaCl₂ to react completely, from stichiometry.

6.2 mol of AgNO₃ need  → ??? mol of CaCl₂ to react completely.

∴ The no. of moles of CaCl₂ are needed = (1 mol)(6.2 mol)/(2 mol) = 3.1 mol.

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4 years ago
What is the diffrence between acidic and basic solutions in working with redox reactions?
Leno4ka [110]
In acidic solutions you have H+ but in basic solutions you have OH-.
You need know that for to balance the reaction.
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3 years ago
Please explain to me how to answer no.2 questions
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Answer: -

IE 1 for X = 801

Here X is told to be in the third period.

So n = 3 for X.

For 1st ionization energy the expression is

IE1 = 13.6 x Z ^2 / n^2

Where Z =atomic number.

Thus Z =( n^2 x IE 1 / 13.6)^(1/2)

Z = ( 3^2 x 801 / 13.6 )^ (1/2)

= 23

Number of electrons = Z = 23

Nearest noble gas = Argon

Argon atomic number = 18

Number of extra electrons = 23 – 18 = 5

a) Electronic Configuration= [Ar] 3d34s2

We know that more the value of atomic radii, lower the force of attraction on the electrons by the nucleus and thus lower the first ionization energy.

So more the first ionization energy, less is the atomic radius.

X has more IE1 than Y.

b) So the atomic radius of X is lesser than that of Y.

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Due to this the remaining electrons are more strongly pulled inside than before ionization. Hence after ionization, the radii of Y decreases.

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4 years ago
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