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Marizza181 [45]
3 years ago
10

Why does the pH of the acid solution initially increase very slowly when metal is first added to the acid solution, but graduall

y increase at a faster rate as the reaction proceeds to completion?
I thought it may be because the metal is just then being completely dissolved but that didn't make sense because we were instructed to wait until the first little scoop of the metal powder was dissolved before adding the next bit.
Chemistry
1 answer:
aleksandr82 [10.1K]3 years ago
5 0
When a metal reacts with an acid, say HCl, it reacts with the H+ atoms of the acid to produce H2 gas and chloride ions. Since pH is the negative log of H+ concentration, then it would sensibly decrease because it was used up. The metal oxide would form a coating over the metal. However, as the reaction proceeds, the acid corrodes the metal and exposes a purer surface removing the coating. Thus increasing the concentration of H+.
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Be sure to answer all parts. What is the [H3O+] and the pH of a buffer that consists of 0.26 M HNO2 and 0.89 M KNO2? (K, of HNO2
Aleksandr-060686 [28]

Answer : The H_3O^+ ion concentration is, 1.12\times 10^{-3}M and the pH of a buffer is, 2.95

Explanation : Given,

K_a=7.1\times 10^{-4}

Concentration of HNO_2 (weak acid)= 0.26 M

Concentration of KNO_2 (conjugate base or salt)= 0.89 M

First we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

pK_a=-\log (7.1\times 10^{-4})

pK_a=4-\log (7.1)

pK_a=3.15

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[KNO_2]}{[HNO_2]}

Now put all the given values in this expression, we get:

pH=3.15+\log (\frac{0.89}{0.26})

pH=2.95

The pH of a buffer is, 2.95

Now we have to calculate the H_3O^+ ion concentration.

pH=-\log [H_3O^+]

2.95=-\log [H_3O^+]

[H_3O^+]=1.12\times 10^{-3}M

The H_3O^+ ion concentration is, 1.12\times 10^{-3}M

4 0
3 years ago
What change in mass number occurs when a radioactive atom emits an alpha particle?
Alik [6]
The alpha particle is the helium nucleus.
The mass of helium nucleus (⁴₂He) is 4u.


The mass of radioactive atom decreases by 4u. Look at the reaction below:

²²³₈₈Ra ---> ⁴₂He + ²¹⁹₈₆Rn
6 0
3 years ago
When the following equation is balanced, what is the coefficient for NaNO3?
notka56 [123]

The coefficient for NaNO₃ = 6

<h3>Further explanation </h3>

Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:  

• 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.  

• 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product  

• 3. Select the coefficient of the substance with the most complex chemical formula equal to 1  

Reaction

AI(NO₃)₃ +Na₂SO₄ → Al₂(SO₄) + NaNO₃

give coefficient

aAI(NO₃)₃ +bNa₂SO₄ → Al₂(SO₄)₃ +c NaNO₃

Al, left=a, right=2⇒a=2

N, left=3a, right=c⇒3a=c⇒3.2=c⇒c=6

Na, left=2b, right=c⇒2b=c⇒2b=6⇒b=3

The equation becomes :

2AI(NO₃)₃ +3Na₂SO₄ → Al₂(SO₄)₃ +6NaNO₃

6 0
3 years ago
If 9.85 grams of copper metal react with 31.0 grams of silver nitrate, how many grams of copper nitrate can be formed and how ma
Veseljchak [2.6K]

Answer:-

29.07 gram of Cu(NO3)2 will be formed.

4.756 grams of AgNO3 will be left over when the reaction is complete.

Explanation:-

Atomic weight of Cu = 63.546 g mol -1

Molecular weight of AgNO3 = 107.87 x 1 + 14 x 1 + 16 x 3

= 169.87 g mol-1

Number of moles of Copper = 9.85 gram / (63.546 g mol-1)

= 0.155 mol

Number of moles of AgNO3 = 31 gram / ( 169.87 g mol-1)

= 0.183 mol

The balanced chemical equation for this reaction is

Cu + AgNO3 --> Cu(NO3)2 + Ag

According to the equation,

1 mole of Cu reacts with 1 mole of AgNO3.

∴0.155 mol of Cu react with 0.155 mol of AgNO3.

Number of moles of AgNO3 left over = 0.183-0.155=0.028 mol

Number of grams of AgNO3 left over = 0.028 mol x 169.87 grams mol-1

= 4.756 gram

Molecular weight of Cu(NO3)2 = 63.546 x 1 + (14 x 1 +16 x 3 ) x 2

=187.546 gram

Now from the balanced chemical equation,

1 Cu gives 1 Cu(NO3)2

∴ 63.546 g of Cu gives 187.546 gram of Cu(NO3)2

9.85 grams pf Cu gives 187.546 x 9.85 / 64.546 gram of Cu(NO3)2

= 29.07 gram of Cu(NO3)2

4 0
3 years ago
Consider the data presented below. time (s) 0 40 80 120 160 moles of a 0.100 0.067 0.045 0.030 0.020 part a part complete determ
Whitepunk [10]
To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows 
time         0         40        80       120       160
moles    0.100   0.067  0.045    0.030    0.020


Q1)
for the first 40 s change of moles ;
      = -d[A] / t
      = - (0.067-0.100)/40s
      = 8.25 x 10⁻⁴ mol/s
for the next 40 s
      = -(0.045-0.067)/40
      = 5.5 x 10⁻⁴ mol/s
the 40 s after that
      = -(0.030-0.045)/40 s 
     = 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction

rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ   -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction

Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation. 
Since this is a first order reaction,
b = 1
therefore the reaction is 
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
   = 8.25 x 10⁻³ s⁻¹

7 0
3 years ago
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