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Dmitriy789 [7]
3 years ago
5

How do I do balancing equations

Chemistry
1 answer:
OlgaM077 [116]3 years ago
7 0

Explanation:

To balance any chemical equation, you must first understand the law of conservation of mass.

The law of conservation of mass states that "atoms are neither created nor destroyed in a chemical reaction but are changed from one form to another".

So in any chemical reaction the number atoms on both sides of the equation must be the same.

    To balance a chemical equation:

  • you can inspect the given equation and fix appropriate whole number coefficient to chemical species that are not balanced.
  • Also you can set up simple and solvable algebraic equations.

For example:

       aNa + bCl₂  →  cNaCl

a, b and c are the coefficients that will balance the equation;

     conserving: Na     a  = c

                          Cl      2b = c

let b = 1

     c = 2

      a = 2

                    2Na + Cl₂  →  2NaCl

learn more

Balanced equation brainly.com/question/2947744

#learnwithBrainly

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sveta [45]

Answer:

E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g)

Explanation:

Which ONE of the following is an oxidation–reduction reaction?

A) PbCO₃(s) + 2 HNO₃(aq) ⇒ Pb(NO₃)₂(aq) + CO₂(g) + H₂O(l). NO. All the elements keep the same oxidation numbers.

B) Na₂O(s) + H₂O(l) ⇒ 2 NaOH(aq). NO. All the elements keep the same oxidation numbers.

C) SO₃(g) + H₂O(l) ⇒ H₂SO₄(aq). NO. All the elements keep the same oxidation numbers.

D) CO₂(g) + H₂O(l) ⇒ H₂CO₃(aq). NO. All the elements keep the same oxidation numbers.

E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g). YES. <u>C is reduced</u> and <u>H is oxidized</u>.

8 0
4 years ago
A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base
yawa3891 [41]

Answer:

11.39

Explanation:

Given that:

pK_{b}=4.82

K_{b}=10^{-4.82}=1.5136\times 10^{-5}

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{1.805\ g}{82.0343\ g/mol}

Moles= 0.022\ moles

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.022}{0.055}

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

                                  B +   H₂O    ⇄     BH⁺ +        OH⁻

At t=0                        0.4                          -              -

At t =equilibrium     (0.4-x)                        x           x            

The expression for dissociation constant is:

K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}

1.5136\times 10^{-5}=\frac {x^2}{0.4-x}

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³  M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>

5 0
3 years ago
How many milliliters of 4.00 M NaOH are required to exactly neutralize 50.0 milliliters of a 2.00 M solution of HNO3 ?
kramer

Answer: The volume of NaOH required is 25.0 ml

Explanation:

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity HNO_3 = 1

M_1 = molarity of HNO_3 solution = 2.00 M

V_1 = volume of  HNO_3 solution = 50.0 ml

n_2 = acidity of NaOH = 1

M_1 = molarity of NaOH solution = 4.00 M

V_1 = volume of  NaOH solution =  ?

Putting in the values we get:

1\times 2.00\times 50.0=1\times 4.00\times V_2

V_2=25.0ml

Therefore, volume of NaOH required is 25.0 ml

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