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3 years ago

Due in a bit! please help tysm btw!

Chemistry

2 answers:

lutik1710 [3]3 years ago

7 0

Unbalanced. Hope this helps!

Free_Kalibri [48]3 years ago

4 0

Answer:

Unbalanced

Explanation:

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WILL GIVE BRAINLIEST

seropon [69]

Answer:

Chemical reaction A, because the reactant is a compound

Explanation:

In a decomposition reaction, a compound is broken down into its components, so the number of products is greater than the number of reactants

3 0

2 years ago

b) 2C2H2(g) + 5O2(g)⟶4CO2(g) + 2H2O(l) 6.54 Calculate the heats of combustion for the following reactions from the standard enth

IRINA_888 [86]

Answer:

a. ΔH⁸ = -1420 kJ/mol b. ΔH⁸ = -1144.84 kJ/mol

Explanation:

C₂H₄ (g) + 3 O₂ (g) ------------------------ 4 CO₂ (g) + 2 H₂O (l) ΔH⁸ = ?

ΔH⁸f kJmol 52.47 0 -399.5 -285.83

ΔH⁸ = 2(-399.5) + 2 (-285.83) - (52.47)

ΔH⁸ = -1420 kJ/mol

H₂S (g) + 3 O₂ (g) ---------------------- 2 H₂O (l) + 2 SO₂ (g)

ΔH⁸f kJmol -20.50 0 -285.83 -296.84

ΔH⁸ = 2(-285.83) + 2 (-296.84) - (-20.50)

ΔH⁸ = -1144.84 kJ/mol

4 0

3 years ago

Which particles that make up an atom are involved in nuclear reactions ?

saveliy_v [14]

Answer: (B) proton And neutrons

Explanation:

8 0

3 years ago

Read 2 more answers

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

Which of the following is least likely to dissolve in the polar solvent water?

s2008m [1.1K]

Like dissolves like and unlike repels like. This applies to the polarity of the solutions. Polar dissolves polar and non-polar does not mix with polar solutions. In this case, the polar substance that can mix with water. Sugar and salt can be dissolved through agitation. Oil which is nonpolar is immiscible in water. Hence answer is D.

6 0

3 years ago