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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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Evgesh-ka [11]

In NH3 , let oxidation number of N be x

x + (+1)3 = 0

x = -3

In HNO3 , let oxidation number of N be x

1 + x + (-2)3 = 0

x = +5

In NO2 , let oxidation number of N be x

x + (-2)2 = 0

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5 0
2 years ago
In a fixed cylinder are 3moles of oxygen gas at 300Kelvin and 1.25atm. What is the volume of the container?
vladimir2022 [97]

Answer:

The volume of the container is 59.112 L

Explanation:

Given that,

Number of moles of Oxygen, n = 3

Temperature of the gas, T = 300 K

Pressure of the gas, P = 1.25 atm

We need to find the volume of the container. For a gas, we know that,

PV = nRT

V is volume

R is gas constant, R =  0.0821 atm-L/mol-K

So,

V=\dfrac{nRT}{P}\\\\V=\dfrac{3\ mol\times 0.0821\ L-atm/mol-K \times 300\ K}{1.25\ atm}\\\\V=59.112\ L

So, the volume of the container is 59.112 L

6 0
2 years ago
The three‑dimensional structure of a generic molecule is given. Identify the axial and equatorial atoms in the three‑dimensional
Dima020 [189]

Answer:

Explanation:

CHECK THE ATTACHMENT FOR THE COMPLETE QUESTION AND THE DETAILED EXPLANATION

NOTE:

Equatorial atoms are referred to atoms that are attached to carbons in the cyclohexane ring which is found at the equator of the ring.

Axial atoms are atoms that exist in a bond which is parallel to the axis of the ring in cyclohexane

4 0
2 years ago
Analyze: The first shell can hold a maximum of two electrons. How does this explain the valence of hydrogen
chubhunter [2.5K]

Answer:

See explanation

Explanation:

Hydrogen has a valency of +1 or -1. Its electronic configuration is 1s1.

The 1s sub-level (first shell) is known to hold two electrons. This means that hydrogen may either loose this one electron in the 1s level to yield H^+ or accept another electron into this 1s level to form H^- (the hydride ion).

The formation of the hydride ion completes the 1s orbital.

4 0
2 years ago
Using the mass of the proton 1.0073 amu and assuming its diameter is 1.0×10−15m, calculate the density of a proton in g/cm3.
icang [17]

Answer : 3.2 X 10^{15} g/cm^{3}

Explanation :  To convert amu i.e. atomic mass unit in grams we have the conversion factor as 1 amu = 1.66054 X 10^{-24} g

we know the mass of the proton is 1.0073 amu

So converting it into grams we have to multiply;

1.0073 amu X  1.66054 X 10^{-24} g/amu = 1.673 X 10^{-24} g

Now, Volume = 1/6πd³ as diameter is given as 1.0 X 10^{-15} m converting it to cm will require to multiply with 100

∴ Volume  = 1/6π (1.0 X 10^{-15}mX 100 cm / 1 m)^{3}

Hence, volume =  5.236 X 10^{-40} cm^{3}

Therefore, Density = mass / volume

∴ Density =  1.673 X 10^{-24} g / 5.236 X 10^{-40} cm^{3}

Therefore, Density will be 3.2 X 10^{15} g/cm^{3}.

6 0
2 years ago
Read 2 more answers
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