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3 years ago

The reaction AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) is a(n) ______________ reaction.

Chemistry

1 answer:

Temka [501]3 years ago

8 0

Answer : The correct option is, (e) precipitation reaction

Explanation :

Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Precipitation reaction : It is defined as the reaction in which an insoluble salt formed when two aqueous solutions are combined.

The insoluble salt that settle down in the solution is known an precipitate.

Single replacement reaction : A chemical reaction in which the more reactive element replace the less reactive element.

It is represented as,

$A+BC\rightarrow AC+B$

In this reaction, A is more reactive element and B is less reactive element.

Acid-base reaction : It is defined as the reaction in which an acid react with a base to give salt and water as a product.

The given chemical reaction is:

$AgNO_3(aq)+NaCl(aq)\rightarrow AgCl(s)+NaNO_3(aq)$

This reaction is a precipitation reaction in which the two aqueous solution silver nitrate and sodium chloride combined to give sodium nitrate solution and silver chloride as a precipitate. In precipitation reaction there is no changes in oxidation state of the element.

Hence, the correct option is, (e) precipitation

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The National Environmental Policy Act was established in 1965. True or false

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The answer to your question is false

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Sewage and industrial pollutants dumped into a body of water can reduce the dissolved oxygen concentration and adversely affect

MArishka [77]

Answer:

FALSE

Since 0.385 < 0.526, the value for week 3 is accepted.

Explanation:

Qexp = (|Xq - Xₙ₋₁|)/w

where Xq is the suspected outlier; Xₙ₋₁ is the next nearest data point; w is the range of data

First, the data are arranged in decreasing order, from highest to lowest:

3. 5.6

2. 5.1

8. 5.1

1. 4.9

6. 4.9

5. 4.7

7. 4.5

4. 4.3

Xq = 5.6; Xₙ₋₁ = 5.1; w = 5.6 - 4.3 = 1.3

Qexp = (|5.6 - 5.1|)/1.3 = 0.385

From tables, at 95% confidence level, for n = 8, Qcrit = 0.526

Since 0.385 < 0.526, the value for week 3 is accepted.

7 0

3 years ago

If equal volumes of 0.1 M HCl and 0.2 M TRIS (base form) are mixed together. The pKa of TRIS is 8.30. Which of the following sta

blondinia [14]

Answer:

option D is correct

D. This solution is a good buffer.

Explanation:

TRIS (HOCH $_{2}$ ) $_{3}$ CNH $_{2}$

if TRIS is react with HCL it will form salt

(HOCH $_{2}$ ) $_{3}$ CNH $_{2}$ + HCL ⇆ (HOCH $_{2}$ ) $_{3}$ NH $_{3}$ CL

Let the reference volume is 100

Mole of TRIS is = 100 × 0.2 = 20

Mole of HCL is = 100 × 0.1 = 10

In the reaction all of the HCL will Consumed,10 moles of the salt will form

and 10 mole of TRIS will left

hence , Final product will be salt +TRIS(9 base)

H = Pk $_{a}$ + log (base/ acid)

8.3 + log(10/10)

8.3

6 0

3 years ago

If 16.29 grams of Na2SO4 is mixed with 3.697 grams of C and allowed to react according to the balanced equation: Na2SO4(aq) + 4

abruzzese [7]

Answer: The limiting reagent in the given chemical reaction is carbon metal.

Explanation:

Excess reagent is defined as the reagent which is present in large amount in a chemical reaction.

Limiting reagent is defined as the reagent which is present in small amount in a chemical reaction. Formation of product depends on the limiting reagent.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For sodium sulfate:

Given mass of sodium sulfate = 16.29 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

$\text{Moles of sodium sulfate}=\frac{16.29g}{142g/mol}=0.115mol$

For carbon:

Given mass of carbon = 3.697 g

Molar mass of carbon = 12 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon}=\frac{3.697g}{12g/mol}=0.31mol$

For the given chemical reaction:

$Na_2SO_4(aq.)+4C(s)\rightarrow Na_2S(s)+4CO(g)$

By Stoichiometry of the reaction:

4 moles of carbon reacts with 1 mole of sodium sulfate

So, 0.31 moles of carbon will react with = $\frac{1}{4}\times 0.31=0.0775mol$ of sodium sulfate

As, given amount of sodium sulfate is more than the required amount. So, it is considered as an excess reagent.

Thus, carbon metal is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reagent in the given chemical reaction is carbon metal.

4 0

3 years ago

Suppose 0.981 g of iron (II) iodide is dissolved in 150. mL of a 35.0 m M aqueous solution of silver nitrate. Calculate the fina

yaroslaw [1]

Answer:

Final molarity of iodide ion C(I-) = 0.0143M

Explanation:

n = (m(FeI(2)))/(M(FeI(2))

Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol

So n = 0.981/309.85 = 0.0031 mol

V(solution) = 150mL = 0.15L

C(AgNO3) = 35mM = 0.035M = 0.035m/L

n(AgNO3) = C(AgNO3) x V(solution)

= 0.035 x 0.15 = 0.00525 mol

(AgNO3) + FeI(3) = AgI(3) + FeNO3

So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol

C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M

8 0

3 years ago