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kogti [31]
3 years ago
9

A laboratory experiment requires 4.8 l of a 2.5 m solution of sulfuric acid (h2so4), but the only available h2so4 is a 6.0 m sto

ck solution. how could you prepare the solution needed for the lab experiment?
Chemistry
2 answers:
Airida [17]3 years ago
5 0
To determine the amount of 6.0 M H2SO4 needed for the preparation, equate the number of moles of the 6.0 M and 2.5 M H2SO4 solution. This is done as follows
 
                                             M1 x V1 = M2 x V2

Substituting the known variables,
 
                                             (6.0 M) x V1 = (2.5 M) x (4.8 L)

Solving for V1 gives an answer of V1 = 2 L. Thus, to prepare the needed solution, dilute 2 L of 6.0 M H2SO4 solution with water until the volume reach 4.8 L. 


Veronika [31]3 years ago
3 0

Answer:

We will take 2 L of sulfuric acid stock solution, and then will add 4L to it.

Explanation:

Given:

Concentration of stock solution = 6 M

volume of stock required = to be calculated =?

Concentration of required solution = 2.5 M

Volume of required solution = 4.8 L

This problem is based on dilution.

We will take some volume of sulfuric acid and will dilute it to 4.8 L.

Now in order to calculate the volume of sulfuric acid required we will use following formula.

M₁V₁=M₂V₂

M₁= molarity of stock solution = 6 M

V₁ = volume of stock solution = ?

M₂ =molarity of required solution = 2.5 M

V₂ = volume of required solution = 4.8 L

Putting values,

V_{1}=\frac{M_{2}V_{2}  }{M_{1} }

V_{1}=\frac{2.5X4.8}{6}=2L

Thus we will take 2 L of sulfuric acid stock solution, and then will add 4L to it.

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Formula for Xenon hexafluoride is XeF6
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When an electron in an atom spontaneously jumps from a higher energy state to a lower energy state, the atom?
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If an atom suffers from a collision, that causes an electron to jump from a lower to higher state, it is called collisional excitation 
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An iron chloride compound contains 55.85 grams of iron and 106.5 grams of chlorine. What is the most likely empirical formula fo
mart [117]

Answer:

FeCl_{3}

Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

mass of Fe = 55.85 g

Molar mass of Fe = 55.85 g/mol

<u>Moles of Fe = 55.85 / 55.85 = 1</u>

mass of Cl = 106.5 g

Molar mass of Cl = 35.5 g/mol

Moles of Cl = 106.5 / 35.5 = 3

Taking the simplest ratio for Fe and Cl as:

1 : 3

The empirical formula is = FeCl_{3}

3 0
3 years ago
How many formula units are there in 212 grams of mgCl2
AleksandrR [38]

Formula units are there in 212 grams of MgCl₂ are 830.56

Formula is the empirical of any ionic or covalent network solid compound used as an independent entity for stoichiometric calculations and it is the lowest whole number ratio of ions represented in an ionic compound

Here given data is

MgCl₂ = 212 grams

1 mole of magnesium chloride has mass = 95.211 gram and contains 6.022×10²³formula units of magnesium chloride

Here 212 grams×6.022×10²³form unit of MgCl₂/95.211 gram = 830.56

Know more about magnesium chloride

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4 0
10 months ago
Neptunium-237 undergoes a series of α-particle and β-particle productions to end up as thallium-205. How many α particles and β
Fantom [35]
<h3>Answer:</h3>

8 alpha particles

4 beta particles

<h3>Explanation:</h3>

<u>We are given;</u>

  • Neptunium-237
  • Thallium-205
  • Neptunium-237 undergoes beta and alpha decay to form Thallium-205.

We are required to determine the number of beta and alpha particles produced to complete the decay series.

  • We need to know that when a radioisotope emits an alpha particle the mass number reduces by 4 while the atomic number decreases by 2.
  • When a beta particle is emitted the mass number of the radioisotope increases by 1 while the atomic number remains the same.

In this case;

Neptunium-237 has an atomic number 93, while,

Thallium-205 has an atomic number 81.

Therefore;

²³⁷₉₃Np → x⁴₂He + y⁰₋₁e + ²⁰⁵₈₁Ti

We can get x and y

237 = 4x + y(0) + 205

237-205 = 4x

4x = 32

 x = 8

On the other hand;

93 = 2x + (-y) + 81

but x = 8

93 = 16 -y + 81

y = 4

Therefore, the complete decay equation is;

²³⁷₉₃Np → 8⁴₂He + 4⁰₋₁e + ²⁰⁵₈₁Ti

Thus, Neptunium-237 emits 8 alpha particles and 4 beta particles to become Thallium-205.

5 0
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