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4 years ago

Which is more prevalent in the food we eat: carbon-12 or carbon-14?

1 answer:

5 0

I think the more prevalent in the food we eat is Carbon-12 which is present in Most of the food we ingest. Both carbon-12 and carbon-14 are isotopes of carbon, they differ in the number of neutrons in each atom. Because of the different number of neutrons, carbon-12 and carbon-14 differ with respect to radioactivity. Carbon-12 is a stable isotope but carbon-14 undergoes radioactive decay.

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A vehicle is traveling at a speed of 92.5 km/h. What is this value in feet per second? Note that, by definition, 1 in = 2.54 cm.

Harlamova29_29 [7]

Answer:

$v=84.3\frac{ft}{s}$

Explanation:

Hello,

In this case, given that 1 kilometer equals 1000 meters, 1 meter equals 3.281 feet and 1 hour equals 3600 seconds, the required velocity in feet per second turns out:

$v=92.5\frac{km}{h}*\frac{1000m}{1km} *\frac{3.281ft}{1m} *\frac{1h}{3600s}\\ \\v=84.3\frac{ft}{s}$

Regards.

7 0

3 years ago

Please Please! help help! so stress

Oksanka [162]

I AM STRESSED AS WELL

8 0

3 years ago

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

Dentify the reagents used to carry out the chlorination of benzene. a. cl2/ccl4 b. cl2/fecl3 c. cl2/nacl d. both b and c e. a, b

algol13

The reagents which are used to carry out the chlorination of benzene are cl2/fecl3 and cl2/nacl

Chlorination is the process of disinfecting drinking water by introducing chlorine to eradicate viruses, germs, and parasites. To get drinking water with safe levels of chlorine, various techniques can be utilized. Chlorination is the technique of disinfecting and eradicating bacteria from drinking water by adding chlorine to it.

To get drinking water with safe levels of chlorine, various techniques can be utilized. Chlorine can be found as solid calcium hypochlorite (Ca(OCl)2), sodium hypochlorite solution (NaOCl), or compressed elemental gas. Chlorine is added to water as part of the chlorination process, however the chlorinating substance need not be pure chlorine.

To learn more about chlorination please visit -
brainly.com/question/14962130
#SPJ4

8 0

2 years ago

3. Explain why an aluminum ion (A13+) is smaller than a sodium ion (Na+).

IceJOKER [234]

Answer:

This question is hard :( I dont know im so sorry

Explanation:

8 0

2 years ago