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timurjin [86]
1 year 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]1 year 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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Determine the number of moles and mass requested for each reaction in Exercise 4.42.
suter [353]

Answer:

(a) 0.22 mol Cl₂ and 15.4g Cl₂

(b) 2.89.10⁻³ mol O₂ and 0.092g O₂

(c) 8 mol NaNO₃ and 680g NaNO₃

(d) 1,666 mol CO₂ and 73,333 g CO₂

(e) 18.87 CuCO₃ and 2,330g CuCO₃

Explanation:

In most stoichiometry problems there are a few steps that we always need to follow.

  1. Step 1: Write the balanced equation
  2. Step 2: Establish the theoretical relationship between the kind of information we have and the one we are looking for. Those relationships can be found in the balanced equation.
  3. Step 3: Apply conversion factor/s to the data provided in the task based on the relationships we found in the previous step.

(a)

Step 1:

2 Na + Cl₂ ⇄ 2 NaCl

Step 2:

In the balanced equation there are 2 moles of Na, thus 2 x 23g = 46g of Na. <u>46g of Na react with 1 mol of Cl₂</u>. Since the molar mass of Cl₂ is 71g/mol, then <u>46g of Na react with 71g of Cl₂</u>.

Step 3:

10.0gNa.\frac{1molCl_{2} }{46gNa} =0.22molCl_{2}

10.0gNa.\frac{71gCl_{2}}{46gNa} =15.4gCl_{2}

(b)

Step 1:

HgO ⇄ Hg + 0.5 O₂

Step 2:

<u>216.5g of HgO</u> form <u>0.5 moles of O₂</u>. <u>216.5g of HgO</u> form <u>16g of O₂</u>.

Step 3:

1.252gHgO.\frac{0.5molO_{2}}{216.5gHgO} =2.89.10^{-3} molO_{2}

1.252gHgO.\frac{16gO_{2}}{216.5gHgO} =0.092gO_{2}

(c)

Step 1:

NaNO₃ ⇄ NaNO₂ + 0.5 O₂

Step 2:

<u>16g of O₂</u> come from <u>1 mol of NaNO₃</u>. <u>16g of O₂</u> come from <u>85g of NaNO₃</u>.

Step 3:

128gO_{2}.\frac{1molNaNO_{3}}{16gO_{2}} =8mol NaNO_{3}

128gO_{2}.\frac{85gNaNO_{3}}{16gO_{2}} =680gNaNO_{3}

(d)

Step 1:

C + O₂ ⇄ CO₂

Step 2:

<u>12 g of C</u> form <u>1 mol of CO₂</u>. <u>12 g of C</u> form <u>44g of CO₂</u>.

Step 3:

20.0kgC.\frac{1,000gC}{1kgC} .\frac{1molCO_{2}}{12gC} =1,666molCO_{2

[tex]20.0kgC.\frac{1,000gC}{1kgC} .\frac{44gCO_{2}}{12gC} =73,333gCO_{2[/tex]

(e)

Step 1:

CuCO₃ ⇄ CuO + CO₂

Step 2:

<u>79.5g of CuO</u> come from <u>1 mol of CuCO₃</u>. <u>79.5g of CuO</u> come from <u>123.5g of CuCO₃</u>.

Step 3:

1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{1mol CuCO_{3}}{79.5gCuO} =18.87molCuCO_{3}\\ 1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{123.5g CuCO_{3}}{79.5gCuO} =2,330gCuCO_{3}

5 0
1 year ago
A molecule contains 23.24 g iodine (I),
gavmur [86]
Convert each amount of grams into moles:

I: 23.24g x 1 mol / 126.90g = 0.1831 mol I

C: 2.198 x 1 mol / 12.01g = 0.1830 mol C

N: 2.562 x 1 mol / 14.01g = 0.1829 mol N

Each element has roughly the same amount of moles, which means the whole number ratio between the elements is 1:1:1

Therefore the empirical formula is ICN
7 0
1 year ago
When a 375 mL sample of nitrogen is kept at constant temperature, it has a pressure of 1.2 atmospheres. What pressure does it ex
kolezko [41]

Answer:

It exerts a pressure of 3.6 atm

Explanation:

This is a gas law problem. We are looking at volume and pressure with temperature being kept constant, thus, the gas law to use is Boyle’s law. It states that at a given constant temperature, the volume of a given mass of gas is inversely proportional to the pressure of the gas.

Mathematically; P1V1 = P2V2

Let’s identify the parameters according to the question.

P1 = 1.2 atm

V1 = 375mL

P2 = ?

v2 = 125mL

We arrange the equation to make room for P2 and this can be written as:

P2 = P1V1/V2

P2 = (1.2 * 375)/125

P2 = 3.6 atm

3 0
1 year ago
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95.8 l of fluorine gas is being held at a temperature of 24.5 c if the temperature were raised to 46.9c what would the new volum
V125BC [204]
V1/T1 = V2/T2
24.5 + 273 = 298.5K
46.9 + 273 = 319.9K

95.8/298 = V2/319.9
V2 = 102.66L
6 0
2 years ago
A beaker contains nickel(ii sulfate, zinc(ii sulfate and solid iron. what species in this beaker will be reduced?
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If a beaker is contained with the substances nickel sulfate, zinc sulfate and solid iron, there is a substance that will be reduced in the following substances that is in it. The substance that will be reduced as it is in the beaker is the Ni^2 + or also known as the Nickel (II)
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1 year ago
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