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Lynna [10]
2 years ago
11

For the neutralization reaction involving hno3 and ca(oh)2, how many liters of 1. 55 m hno3 are needed to react with 45. 8 ml of

a 4. 66 m ca(oh)2 solution?
Chemistry
1 answer:
Agata [3.3K]2 years ago
6 0

For the neutralization reaction involving HNO₃ and Ca(OH)₂, liters of 1.55 M HNO₃ are needed to react with 45.8 ml of a 4.66 m Ca(OH)₂ solution is 0.270 L.

<h3>What is neutralization reaction?</h3>

Those reactions in which acids and bases will combine with each other for the formation of water molecule and salt, is known as neutralization reaction.

Given chemical reaction is:

2HNO₃ + Ca(OH)₂ → 2H₂O + Ca(NO₃)₂

Moles (n) of Ca(OH)₂ will be calculated by using the below formula:
M = n/V, where

M = molarity = 4.66M

V = volume = 45.8 mL = 0.045 L

n = (4.66)(0.045) = 0.2097 mol

From the stoichiometry of the reaction,

1 mole of Ca(OH)₂ = reacts with 2 moles of HNO₃

0.2097 mole of Ca(OH)₂ = reacts with 2×0.2097=0.419 moles of HNO₃

Again by using the molarity equation volume will be calculated for HNO₃ as:

V = (0.419) / (1.55) = 0.270 L

Hence required volume of HNO₃ is 0.270 L.

To know more about neutralization reaction, visit the below link:

brainly.com/question/23008798

#SPJ4

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Determine the change in volume that takes place when a 2.00 L sample of N2 gas is heated from 250 C to 500 C.
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Answer:

V₂ = 2.96 L

Explanation:

Given data:

Initial volume = 2.00 L

Initial temperature =  250°C

Final volume = ?

Final temperature = 500°C

Solution:

First of all we will convert the temperature into kelvin.

250+273 = 523 k

500+273= 773 k

According to Charles's law,

V∝ T

V = KT

V₁/T₁ = V₂/T₂

V₂ = T₂V₁/T₁

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Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

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 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

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