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kolbaska11 [484]
2 years ago
15

A student is curious about the Ksp value for NaCl. The student looks up the value om the appendix of his textbook but cannot fin

d a value for NaCl. Why not?
Chemistry
1 answer:
Leno4ka [110]2 years ago
7 0

Given :

The student looks up the value of the appendix of his textbook but cannot find a value for NaCl.

To Find :

Why the Ksp of NaCl is not present .

Solution :

Ksp values are given for those compounds which are partially soluble or insoluble in water .

But , we know NaCl is strong electrolyte and it completely dissociated in water .

So , the reaction go forward and their is no need to calculate Ksp .

Therefore , the Ksp of NaCl is not present .

Hence , this is the required solution .

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Which of the following statements is (are) true about enzyme-catalyzed reactions? a. The reaction is faster than the same reacti
Brums [2.3K]

Answer : The correct option is A.

Explanation :

Enzyme-catalyzed reaction :

Enzyme act as a biological catalyst and the role of catalyst is to increase the rate of chemical reaction by lowering the activation energy.

Most of the chemical reactions are slow in the absence of enzyme but in the presence of enzyme, the reaction become faster. That means the Enzyme accelerate the rate of reaction.

Therefore, the correct answer is the reaction is faster than the same reaction in the absence of the enzyme.

5 0
2 years ago
HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.
Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, C_A = 6.0mol/dm³

volume of HA, V_A  = 25.0cm³, = 0.025dm³

Concentration of HB, C_B = 3.0mol/dm³

volume of HB, V_B = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

C_iVi = C_fV_f

where;

C_i is initial concentration

V_i is initial volume

C_f is final concentration of the solution

V_f is final volume of the solution

C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

7 0
2 years ago
1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl
kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

3 0
2 years ago
When balanced, which equation would have the coefficient 3 in front of any of the reactants? Zn + HCl ZnCl2 + H2 H2SO4 + B(OH)3
Gwar [14]

The correct answer is B. H2SO4 + B(OH)3  B2(SO4)3 + H2O

Hope this helps!

6 0
2 years ago
Read 2 more answers
Determine the number of bonding electrons and the number of nonbonding electrons in the structure of xef2. enter the number of b
lakkis [162]

Answer : The number of bonding electrons and the number of non-bonding electrons are (4, 18).

Explanation :

The number of bonding electrons and non-bonding electrons in the structure of XeF_2 is determined by the Lewis-dot structure.

Lewis-dot structure : It tell us about the number of valence electrons of an atom within a molecule and it is also shows the bonding between the atoms of a molecule and the lone-pair of electrons.

In the given structure, 'Xe' is the central atom and 'F' is the terminal atom.

Xenon has 8 valence electrons and fluorine has 7 valence electrons.

Total number of valence electrons in XeF_2 = 8 + 2(7) = 22 electrons

From the Lewis-dot structure, we conclude that

The number of electrons used in bonding = 4

The number of electrons used in non-bonding (lone-pairs) = 22 - 4 = 18

Therefore, the number of bonding electrons and the number of non-bonding electrons are (4, 18).

The Lewis-dot structure of XeF_2 is shown below.

4 0
2 years ago
Read 2 more answers
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