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3 years ago

In titrating 0.20 M hydrochloric acid, HCl, with 0.20 M NaOH at 25°C, the solution at the equivalence point is

Chemistry

1 answer:

xxTIMURxx [149]3 years ago

8 0

Answer:

c) 0.10M NaCl

Explanation:

<em>Options are:</em>

<em>b) Very acidic</em>

<em>e) Slightly acidic.</em>

<em />

Based on the reaction:

NaOH + HCl → H₂O + NaCl

You can see than 1 mole of NaOH reacts per mole of HCl producing 1 mole of NaCl

Assuming 1L of 0.20M HCl reacts with 1L of NaOH, moles of both compounds are:

1L * 0.20mol/L = 0.20 moles reacts. That means moles of NaCl produced are 0.20moles in 2L. Molarity of NaCl is:

0.20mol / 2L = 0.10M NaCl

That means, right option is:

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Give two reasons why a luminous flame is not used for heating purposes. (2mks) a

Natali [406]

Answer:

The Two reasons are :-

1. Combustion efficiency

2. Standardization of test conditions as non-luminous flame.

Explanation:

A luminous flame isn't suitable for heating as it gives out soot (A black powdery or flaky substance consisting largely of amorphous carbon, produced by the incomplete burning of organic matter)

8 0

3 years ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

<u>Given:</u>

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

<u>Given:</u>

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

5 0

4 years ago

If 2.0 g of copper(II) chloride react with excess sodium nitrate, what mass of sodium chloride is formed in this double replacem

cluponka [151]

Taking into account the reaction stoichiometry, 1.729 grams of NaCl is formed.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CuCl₂ + 2 NaNO₃ → Cu(NO₃)₂ + 2 NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CuCl₂: 1 mole
NaNO₃: 2 moles
Cu(NO₃)₂ : 1 mole
NaCl: 2 moles

The molar mass of the compounds is:

CuCl₂: 134.44 g/mole
NaNO₃: 85 g/mole
Cu(NO₃)₂ : 187.54 g/mole
NaCl: 58.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

CuCl₂: 1 mole ×134.44 g/mole= 134.44 grams
NaNO₃: 2 moles ×85 g/mole= 170 grams
Cu(NO₃)₂ : 1 mole ×187.54 g/mole= 187.54 grams
NaCl: 2 moles ×58.45 g/mole= 116.9 grams

<h3>Mass of NaCl formed</h3>

The following rule of three can be applied: If by stoichiometry of the reaction 134.44 grams of CuCl₂ form 116.9 grams of NaCl, 2 grams of CuCl₂ form how much mass of NaCl?

$mass of NaCl=\frac{2 grams of CuCl_{2}x116.9 grams of NaCl }{134.44grams of CuCl_{2}}$

<u><em>mass of NaCl= 1.739 grams</em></u>

Finally, 1.729 grams of NaCl is formed.

Learn more about the reaction stoichiometry:

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algol13

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