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olga nikolaevna [1]
3 years ago
9

I’ll give you THE BRAINLIEST

Chemistry
1 answer:
Oxana [17]3 years ago
5 0

hope you understand everything , (AD).

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How many milliliters of a 0.211 M HI solution are needed to reduce 24.0 mL of a 0.354 M KMnO4 solution according to the followin
Novay_Z [31]

Answer:

The answer to your question is 242 ml

Explanation:

Data

HI 0.211 M   Volume = x

KMnO₄ 0.354 M   Volume = 24 ml

Balanced Chemical reaction

     12HI + 2KMnO₄ + 2H₂SO₄ → 6I₂ + Mn₂SO₄ + K₂SO₄ + 8H₂O

Process

1.- Calculate the moles of KMnO₄  0.354 M in 24 ml

Molarity = moles / volume (L)

moles = Molarity x volume (L)

moles = 0.354 x 0.024

moles = 0.0085

2.- From the balanced chemical reaction we know that HI and KMnO₄ react in the proportion 12 to 2. Then,

              12 moles of HI --------------- 2 moles of KMnO₄

                x                     --------------- 0.0085 moles of KMnO₄

             x = (0.0085 x 12)/2

             x = 0.051 moles of HI

3.- Calculate the milliliters of HI 0.211 M

Molarity = moles/volume

Volume = moles/molarity

Volume = 0.051/0.211

Volume = 0.242 L or Volume = 242 ml

8 0
3 years ago
Identify the balanced equation for the following reaction:<br><br> SO2(g) + O2(g) → SO3(g)
sergiy2304 [10]

Answer:  The balanced equation for the given reaction is

2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g).

Explanation:

A chemical equation which contains same number of atoms on both reactant and product side.

For example, SO_{2}(g) + O_{2}(g) \rightarrow SO_{3}(g)

Here, number of atoms on reactant side are as follows.

  • S = 1
  • O = 4

Number of atoms on product side are as follows.

  • S = 1
  • O = 3

To balance this equation, multiply SO_{2} by 2 on reactant side and multiply SO_{3} by 2. Hence, the equation will be re-written as follows.

2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)

Here, number of atoms on reactant side are as follows.

  • S = 2
  • O = 6

Number of atoms on product side are as follows.

  • S = 2
  • O = 6

Now, there are same number of atoms on both reactant and product side. So, this equation is balanced.

Thus, we can conclude that the balanced equation for the given reaction is 2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g).

3 0
3 years ago
Which statement best describes a solution?
lara31 [8.8K]

A mixture having a uniform composition where the components can't be seen separately and all components are in the same state best describes a solution.

 

<span>In chemistry, a </span>solution<span> is a homogeneous mixture composed of two or more substances. In such a mixture, a solute is a substance dissolved in another substance, known as a solvent.</span>

 

The correct answer between all the choices given is the third choice or letter C. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

6 0
3 years ago
Read 2 more answers
A buffer is composed of nh3 and nh4cl. How would this buffer solution control the ph of a solution when a small amount of a stro
Shkiper50 [21]

Answer:

According to Le-chatelier principle, equilibrium will shift towards left to minimize concentration of OH^{-} and keep same equilibrium constant

Explanation:

In this buffer following equilibrium exists -

NH_{3}(aq.)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq.)+OH^{-}(aq.)

So, OH^{-} is involved in the above equilibrium.

When a strong base is added to this buffer, then concentration of OH^{-} increases. Hence, according to Le-chatelier principle, above equilibrium will shift towards left to minimize concentration of OH^{-} and keep same equilibrium constant.

Therefore excess amount of OH^{-} combines with NH_{4}^{+} to produce ammonia and water. So, effect of addition of strong base on pH of buffer gets minimized.

5 0
3 years ago
A solution is made containing 4.6 g of sodium chloride per 250g of water.
lozanna [386]

Answer:

\large \boxed{1.81 \, \%}

Explanation:

\text{Mass percent} = \dfrac{\text{mass of solute} }{\text{mass of solution}} \times 100 \, \%

Data:

Mass of NaCl =      4.6 g

Mass of water = 250    g

Calculations:

Mass of solution = mass of NaCl + mass of water = 4.6 g + 250 g = 254.6 g.

\text{\% m/m} = \dfrac{\text{4.6 g} }{\text{254.6 g}} \times 100 \, \% = \mathbf{1.81 \, \%}\\\\\text{The percent by mass of NaCl is $\large \boxed{\mathbf{1.81 \, \%}}$}

3 0
3 years ago
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