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timurjin [86]
2 years ago
6

g You observed the formation of several precipitates in the Reactions in Solution lab exercise. Identify the precipitate in each

of the following reactions: a. The yellow precipitate formed in the reaction between KI and Pb(NO3)2 is . b. The white precipitate formed in the reaction between BaCl2 and H2SO4 is . c. The brown precipitate formed in the reaction between NaOH and FeCl3 is . d. The blue precipitate formed in the reaction between CuSO4 and NaOH is .
Chemistry
1 answer:
RUDIKE [14]2 years ago
6 0

<u>Answer:</u>

<u>For a:</u> Lead iodide is a yellow precipitate.

<u>For b:</u> Barium sulfate is a white precipitate.

<u>For c:</u> Ferric hydroxide is a brown precipitate.

<u>For d:</u> Copper (II) hydroxide is a blue precipitate.

<u>Explanation:</u>

Precipitation reaction is defined as the reaction where a solid precipitate (solid substance) is formed at the end of the reaction. It is insoluble in water.

For the given options:

  • <u>For (a):</u>

The chemical reaction between KI and lead (II) nitrate follows:

2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)

The iodide of lead is generally insoluble in water. Thus, lead iodide is a yellow precipitate.

  • <u>For b:</u>

The chemical reaction between barium chloride and sulfuric acid follows:

BaCl_2(aq)+H_2SO_4(aq)\rightarrow BaSO_4(s)+2HCl(aq)

The sulfate of barium is insoluble in water. Thus, barium sulfate is a white precipitate.

  • <u>For c:</u>

The chemical reaction between NaOH and ferric chloride follows:

3NaOH(aq)+FeCl_3(aq)\rightarrow Fe(OH)_3(s)+3NaCl(aq)

The hydroxide of iron is insoluble in water. Thus, ferric hydroxide is a brown precipitate.

  • <u>For d:</u>

The chemical reaction between NaOH and copper sulfate follows:

CuSO_4+2NaOH\rightarrow Cu(OH)_2+Na_2SO_4

The hydroxide of copper is insoluble in water. Thus, copper (II) hydroxide is a blue precipitate.

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The following three equations represent reactions done in this lab exercise. Classify each reaction by their type. Note there ma
gulaghasi [49]

Answer: 1. Cu(s)+4HNO_3(aq)rightarrow Cu(NO_3)_2(aq)+2NO_2(g)+2H_2O(l) : oxidation reduction

2. Cu(NO_3)_2(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+2NaNO_3(aq) : precipitation

3. CuO(s)+2HCl(aq)\rightarrow CuCl_2(aq)+H_2O(l) : Double displacement

Explanation:

Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

Cu(s)+4HNO_3(aq)rightarrow Cu(NO_3)_2(aq)+2NO_2(g)+2H_2O(l)

Double displacement reaction is defined as the reaction where exchange of ions takes place. Double displacement reaction in which one of the product remain in solid form are represented by (s) after their chemical formulas. Such double displacement reaction are called as precipitation reaction.

Cu(NO_3)_2(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+2NaNO_3(aq)

Double displacement reaction is defined as the reaction where exchange of ions takes place.

CuO(s)+2HCl(aq)\rightarrow CuCl_2(aq)+H_2O(l)

Single displacement reaction is defined as the reaction where more reactive element displaces a less reactive element from its chemical reaction.

Decomposition reaction is defined as the reaction where a single substance breaks down into two or more simpler substances.

Synthesis/Combination reaction is defined as the reaction where substances combine in their elemental state to form a single compound.

6 0
3 years ago
When you convert a measurement from SI to English, what changes? units , values , error , or resolution
Julli [10]
The answer to your question is units and values
6 0
3 years ago
If the pressure on a gas at -73°C is doubled but its volume is held constant, what will its final temperature be in degrees Cels
Gre4nikov [31]

Answer:

127°C

Explanation:

This excersise can be solved, with the Charles Gay Lussac law, where the pressure of the gas is modified according to absolute T°.

We convert our value to K → -73°C + 273 = 200 K

The moles are the same, and the volume is also the same:

P₁ / T₁ = P₂ / T₂

But the pressure is doubled so: P₁ / T₁ = 2P₁ / T₂

P₁ / 200K = 2P₁ / T₂

1 /2OOK = (2P₁ / T₂) / P₁

See how's P₁ term is cancelled.

200K⁻¹ = 2/ T₂

T₂ = 2 / 200K⁻¹  → 400K

We convert the T° to C → 400 K - 273 = 127°C

4 0
3 years ago
5. An object is a regular, rectangular, solid with dimensions of 2 cm by 3cm by 2cm. It has a mass of 24 g. find its density.
m_a_m_a [10]

Answer:

Density, D = 2g/cm^3

Explanation:

Given the following data;

Length = 2cm

Width = 3cm

Height = 2cm

Mass = 24g

Density = ?

Volume of a rectangular solid (V) = Length × Weight × Height

Therefore, V = L× W × H

Substituting the values, we have;

V = 2 *3 * 2 = 12

V = 12cm

Density can be defined as the ratio of mass to volume i.e mass all over volume.

Mathematically, Density = \frac{Mass}{Volume}

D = \frac{M}{V}

Substituting the values, we have;

D = \frac{24}{12}

<em>Density, D = 2g/cm^3</em>

Hence, the density of the rectangular solid is 2g/cm^3.

3 0
3 years ago
The PH of a 0.1 M MCl (M is an unknown cation) was found to be 4.7. Write the net ionic equation for the hydrolysis of M and its
Serggg [28]

Answer:

6.25 X10^{-9} = Ka

Kb= \frac{10^{-14}}{6.25 X10^{-9}}= 1.6X10^{-6}

Explanation:

The ionic equation for the hydrolysis of the cation of the given salt will be:

M^{+} + H_{2}O ---> MOH + H^{+}

The expression for Ka will be:

Ka = \frac{[H^{+}][MOH]}{[M^{+}]}

As given that the concentration of the salt is 0.1 M and pH of solution = 4.7, we can determine the concentration of Hydrogen ions from the pH

pH = -log [H⁺]

[H⁺] = antilog(-pH) = antilog (-4.7) = 2 X 10⁻⁵ M = [MOH]

Let us calculate Ka from this,

Ka = {2.5X10^{-5}X2.5X10^{-5}}{0.1} = 6.25 X10^{-9}

The relation between Ka an Kb is

KaXKb =10⁻¹⁴

Kb= \frac{10^{-14}}{6.25 X10^{-9}}= 1.6X10^{-6}

6 0
3 years ago
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