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3 years ago

Why would heating up the human body be a bad choice for causing a chemical reaction?

Chemistry

1 answer:

Furkat [3]3 years ago

5 0

Answer:

Explained below

Explanation:

The human body has a normal core temperature of around 37°C to 38°C.

Now, if it is heated up to say 39° to 40°C, fatigue will start to set in and the brain begins to tell the muscles to slow down.

If it's now heated to higher temperatures above above 41°C, the body will begin to experience heat exhaustion and therefore will start to shut down.

Due to this process, the body can't even sweat at that stage because blood flow to the skin will stops thereby making the body feel cold and clammy. Thus, chemical processes/reaction in the body will begin to be affected and the cells inside the body will start to deteriorate and thus there is now a huge risk of having multiple organ failure.

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Balance the following redox equations by the half-reaction method:

frozen [14]

Answer:

a) Mn^2+ + 2OH- + H2O2 → MnO2 + 2H2O

b) 2Bi(OH)3 + 3SnO2^2- → 2Bi + 3SnO3^2- + 3H2O

c) Cr2O7^2- + 3C2O4^2- + 14H+ → 2Cr^3+ +7H2O + 6CO2

d) 2ClO3- + 2Cl- + 4H+ → Cl2 + 2H2O +2ClO2

e) 5 BiO3^- + 14 H+ + 2Mn^2+ → 5Bi^3+ + 7H2O + 2MnO4^-

Explanation:

(a) Mn2 + H2O2 → MnO2 + H2O (in basic solution)

Step 1: The half reactions

Oxidation: Mn2+ + 4OH- → MnO2 + 2H2O + 2e-

Reduction: H2O2 + 2e- + 2H2O → 2H2O + 2OH-

Step 2: Sum of both half reactions

Mn2+ + 4OH- + H2O2 → MnO2 + 2H2O + 2OH-

Step 3: the netto reaction

Mn^2+ + 2OH- + H2O2 → MnO2 + 2H2O

(b) Bi(OH)3 + SnO2^2- → SnO3^2- + Bi (in basic solution)

Step 1: The half reactions

Reduction: Bi(OH)3 + 3e- → Bi

Oxidation : Sno2^2- → SnO3^2- +2e-

Step 2: Balance the half reactions

2* (Bi(OH)3 + 3e- → Bi + 3OH-)

3* (Sno2^2- +2OH- → SnO3^2- +2e- + H2O)

Step 3: The netto reaction

2Bi(OH)3 + 3SnO2^2- → 2Bi + 3SnO3^2- + 3H2O

(c) Cr2O7^2- + C2O4^2- → Cr^3+ + CO2 (in acidic solution)

Step 1: The half reactions

Reduction: Cr2O7^2- + 6e- → 2Cr+

Oxidation : C2O4^2- → 2CO2 + 2e-

Step 2: Balance the half reactions

Cr2O7^2- + 6e- +14H+ → 2Cr+ +7H2O

3*(C2O4^2- → 2CO2 + 2e-)

Step 3: The netto reaction

Cr2O7^2- + 3C2O4^2- + 14H+ → 2Cr^3+ +7H2O + 6CO2

(d) ClO3^- + Cl^− → Cl^2 + ClO^2 (in acidic solution)

Step 1: The half reactions

Reduction: 2 ClO3^- + 10e- → Cl2

ClO3^- + e- → ClO2

2 Cl- + 2ClO3^- +8e- →2Cl2

Oxidation: 2Cl- → Cl2 + 2e-

Cl- → ClO2 + 5e-

Cl- +ClO3^- → 2ClO2 + 4e-

Step 2: Balance the reactions

2Cl- + 2ClO3^- + 8e- + 12H+ → 2Cl2 + 6H2O

2* (Cl- + ClO3^- + H2O → 2ClO2 + 4e- + 2 H+)

Step 3: The netto reaction

2ClO3^- + 2Cl- + 4H+ → Cl2 + 2H2O +2ClO2

(e) Mn^2 + BiO3^− → Bi^3 + MnO^4− (in acidic solution)

Step 1: The half reactions

Reduction: BiO3^- + 2e- → Bi^3+

Oxidation : Mn^2+ → MnO4^- +5e-

Step 2: Balanced the reactions

5* ( BiO3^- + 2e- + 6H+ → Bi^3+ + 3H2O)

2* ( Mn^2+ + 4H2O →MnO4^- + 5e- + 8H+)

Step 3: The netto reaction

5 BiO3^- + 14 H+ + 2Mn^2+ → 5Bi^3+ + 7H2O + 2MnO4^-

6 0

4 years ago

What is the percent yield for the reaction below when

kumpel [21]

Answer:

Percent yield of reaction is 150%.

Explanation:

Given data:

Percent yield = ?

Actual yield of SO₃ = 586.0 g

Mass of SO₂ = 705.0 g

Mass of O₂ = 80.0 g

Solution:

Chemical equation:

2SO₂ + O₂ → 2SO₃

Number of moles of SO₂:

Number of moles = mass/ molar mass

Number of moles = 586.0 g/ 64.1 g/mol

Number of moles = 9.1 mol

Number of moles of O₂:

Number of moles = mass/ molar mass

Number of moles = 80.0 g/ 32g/mol

Number of moles = 2.5 mol

Now we will compare the mole of SO₃ with O₂ and SO₂.

SO₂ : SO₃

2 : 2

9.1 : 9.1

O₂ : SO₃

1 : 2

2.5 : 2×2.5 = 5

The number of moles of SO₃ produced by oxygen are less it will limiting reactant.

Theoretical yield of SO₃:

Mass = number of moles × molar mass

Mass = 5 mol × 80.1 g/mol

Mass = 400.5 g

Percent yield of reaction:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 586.0 g/ 400.5 g× 100

Percent yield = 1.5× 100

Percent yield = 150%

8 0

3 years ago

What product(s) (excluding stereoisomers) is/are formed when y is heated with br2?

KatRina [158]

When the molecule undergoes chlorination with Cl2 on heating, the hydrogen atom of the alkyl group is replaced by the chlorine atom and form chloroalkanes. The molecule X containes three types of alkyl halides, therefore three different types of chloroalkanes are formed by the replacement of hydrogem atom linked to these alkyl groups. Thus the three different types of chloroalkanes are formed - primary, secondary, and thertiary chloroalkane. Chlorination is not selective so a mixture pf products results. The products formed by the reaction of the molecule with Y with Cl2 are shown on the attached file.