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Kay [80]
2 years ago
14

List the type of reaction. Predict the outcome and balance.

Chemistry
1 answer:
Yuliya22 [10]2 years ago
3 0

Answer:

Single Displacement Reaction: a reaction in which an element is swapped for another in a molecule

A + BC ---> AC + B

Double Displacement Reaction: a reaction in which the cation of one molecule is swapped with a cation of another molecule

AB + CD ---> AD + CB

a) Single Displacement Reaction

3 Ca + 2 AlBr₃ ----> 3 CaBr₂ + 2 Al

b) Single Displacement Reaction

2 Al + 3 MgSO₄ ----> Al₂(SO₄)₃ + 3 Mg

c) Double Displacement Reaction

AgNO₃ + KCl ----> AgCl + KNO₃

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Nat2105 [25]

A good reason for a desert fox to show this pattern of behavior because hunting at night allows the fox to use its night vision.

<h3>What is Hunting?</h3>

Thi9s is commonly practised by predators such as fox in which they capture and kill other animals for food.

The fox has a good night vision which makes it able to hunt for animals during the night also. This is why option C is chosen as the most appropriate choice.

Read more about Hunting here  brainly.com/question/81175

6 0
2 years ago
Which statement is correct? A. pKa is not an indicator of acid strength. B. An acid with a small Ka is stronger than an acid wit
Svetradugi [14.3K]

Answer:

D

Explanation:

Since [pKa = - log Ka]....hence..,the larger the Ka value,the stronger the acid is..so this means that the pKa is vice versa

Saying that the smaller the pKa value..the stronger the acid is.

3 0
3 years ago
Read 2 more answers
What is the theoretical yield of fluorenone if you oxidize 175 mg of fluorene?
My name is Ann [436]

Answer:

  • 602 mg of CO₂ and 94.8 mg of H₂O

Explanation:

The<em> yield</em> is measured by the amount of each product produced by the reaction.

The chemical formula of <em>fluorene</em> is C₁₃H₁₀, and its molar mass is 166.223 g/mol.

The <em>oxidation</em>, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

        2C_{13}H_{10}+31O_2\rightarrow 26CO_2+10H_2O

To calculate the yield follow these steps:

<u>1. Mole ratio</u>

          2molC_{13}H_{10}:31molO_2:26molCO_2:10molH_2O

<u />

<u>2. Convert 175mg of fluorene to number of moles</u>

  • 175mg/times 1g/1,000mg=0.175g

  • Number of moles = mass in grams / molar mass

  • \text{number of moles}=0.175g/166.223g/mol=0.0010528mol

<u>3. Set a proportion for each product of the reaction</u>

a) <u>For CO₂</u>

i) number of moles

         2molC_{10}H_{13}/26molCO_2=0.0010528molC_{10}H{13}/x

x=0.0010528molC_{10}H_{13}\times 26molCO_2/2molC_{10}H_{13}=0.013686molCO_2

ii) mass in grams

The molar mass of CO₂ is 44.01g/mol

  • mass = number of moles × molar mass
  • mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg

b) <u>For H₂O</u>

i) number of moles

0.0010528molC_{10}H_{13}\times10molH_2O/2molC_{10}H_{13}=0.00526molH_2O

ii) mass in grams

The molar mass of H₂O is 18.015g/mol

  • mass = number of moles × molar mass
  • mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg
4 0
3 years ago
What is the percent by mass of carbon in C10 H14 N2?
MrMuchimi
Multiply the number of each element by its mass on the periodic table.

Add them together.

Put the mass of carbon only over the total mass and multiply the result by 100 to get the percentage.
8 0
3 years ago
Read 2 more answers
Find the potentials of the following electrochemical cell:
melomori [17]

Answer: 0.18 V

Explanation:-

Cd/Cd^{2+}(0.10M)//Ni^{2+}(0.50M)?Ni

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

E^0_(Cd^{2+}/Cd)=-0.40V[/tex]

E^0_(Ni^{2+}/Ni)=-0.24V[/tex]

Cd+Ni^{2+}\rightarrow Cd^{2+}+Ni

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials.

E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Cd^{2+}/Cd]}

E^0=-0.24-(-0.40)=0.16V

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cd^{2+}]}{[Ni^{2+]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential = 0.16 V

E_{cell}=0.16-\frac{0.0592}{2}\log \frac{[0.10]}{[0.5]}

E_{cell}=0.18V

Thus the potential of the following electrochemical cell is 0.18 V.

6 0
3 years ago
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