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Galina-37 [17]
1 year ago
10

Question is in picture below

Chemistry
1 answer:
liq [111]1 year ago
8 0

The dissociation of cadmium chloride is as follows: CdCl₂(s) → Cd⁺²(aq) + 2Cl⁻(aq)

<h3>What is dissociation?</h3>

Dissociation is the process by which a compound body breaks up into simpler constituents; said particularly of the action of heat on gaseous or volatile substances.

It is a chemical reaction in which a compound breaks apart into two or more components. The general formula for a dissociation reaction follows the form:

AB → A + B

According to this question, cadmium chloride undergoes dissociation into cadmium and chlorine ions as follows:

CdCl₂(s) → Cd⁺²(aq) + 2Cl⁻(aq)

Learn more about dissociation at: brainly.com/question/28952043

#SPJ1

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Assuming your average walking speed is 1.4 m/s, what time should you leave your house to arrive at school (0.84 km away) at 7:15
Juli2301 [7.4K]

Answer: 6:50

Explanation:

I think that is correct is use the math correctly u could get 6:50 like i got or 7:00.

3 0
3 years ago
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It has been hypothesized that a chemical known as BW prevents colds. To test this hypothesis, 20,000 volunteers were divided int
Amiraneli [1.4K]

Grams of BW

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8 0
2 years ago
When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for
ruslelena [56]

Answer:

b. 16, reactant side

Explanation:

Let's consider the following redox reaction.

MnO₄⁻(aq) + I⁻(aq) → Mn²⁺(aq) + I₂(s)

We can balance it using the ion-electron method.

Step 1: Identify both half-reactions

Reduction: MnO₄⁻(aq) → Mn²⁺(aq)

Oxidation: I⁻(aq) → I₂(s)

Step 2: Perform the mass balance, adding H⁺(aq) and H₂O(l) where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)

Step 3: Perform the charge balance, adding electrons where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)  + 2 e⁻

Step 4: Multiply both half-reactions by numbers so that the number of electrons gained and lost are equal

2 × (MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 I⁻(aq) → I₂(s)  + 2 e⁻)

Step 5: Add both half-reactions and cancel what is repeated on both sides

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 e⁻ + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)  + 10 e⁻

The balanced reaction is:

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)

5 0
2 years ago
Help please!
Katyanochek1 [597]

The observation, in this case, is quantitative.

<h3>Quantitative observation</h3>

Quantitative observations are observations that can be recorded based on quantitative data. In other words, they are observations that can be assigned numerical values.

Quantitative observations are as opposed to qualitative observations because the former cannot be assigned numerical values. They can be ranked or qualified.

In this case, Danielle and Heather could assign numbers to the length, width, and height of the tank in order to calculate its volume.

More on quantitative observations can be found here: brainly.com/question/17491501

3 0
2 years ago
An excess of oxygen reacts with 451.4 g of lead, forming 374.7 g of lead(II) oxide. Calculate the percent yield of the reaction.
Stels [109]

Answer: The percent yield of the reaction is 77.0 %

Explanation:

2Pb+O_2\rightarrow 2PbO

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

\text{Number of moles of lead}=\frac{451.4g}{207.2g/mol}=2.18moles

\text{Number of moles of lead oxide}=\frac{374.7g}{223.2g/mol}=1.68moles

According to stoichiometry:

2 moles of Pb produces = 2 moles of PbO_2

2.18 moles of Pb is produced by=\frac{2}{2}\times 2.18=2.18moles of PbO_2

Mass of PbO_2 =moles\times {\text {Molar mass}}=2.18\times 223.2g/mol=486.6

percent yield =\frac{374.7g}{486.6g}\times 100=77.0\%

3 0
3 years ago
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