All categories

1 year ago

The American Heart Association recommends to eat no more than 2,301mg of sodium per day. Convert the mass of

Chemistry

1 answer:

castortr0y [4]1 year ago

4 0

The mass of 2,301 grams of sodium in ounces is 0.0811757609 ounces.

The amount of sodium recommended by American heart association is 2301 mg. This limit should not be crossed in a day.

We have to convert Mass of sodium from mg to ounces.

We know,

1 ounce = 28.3459 grams.

We also know,

1 gram = 1000 milligrams.

So,

28.3459 grams = 28.3459 x 1000 milligrams.

28.3459 grams = 28345.9 milligrams.

1 ounce = 28345.9 mg.

1 mg = 1/28345.9 ounces

Weight of sodium 2301mg in ounces,

2301mg = 2301/28345.9 ounces

Dividing till last significant figure,

2301 mg = 0.0811757609 ounces.

Mass of sodium in ounces is 0.0811757609.

To know more about Unit conversion, visit,

brainly.com/question/97386

#SPJ9

You might be interested in

Which element is likely to have the highest thermal conductivity

givi [52]

sodium element is likely to have the highest thermal conductivity

7 0

3 years ago

The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:

german

Answer:

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

Explanation:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

We are given:

$Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Cell having 1st and 2nd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-0.337=0.462V$

Cell having 1st and 3rd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.28)=1.079V$

Cell having 1st and 4th half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.74)=1.539V$

Cell having 2nd and 3rd half reactions:

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.337-(-0.28)=0.617V$

Cell having 3rd and 4th half reactions:

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=-0.28-(-0.74)=0.46V$