How many grams of nitrogen are needed to produce 812 grams of ammonia?

When ethanol is burned, it's converted into carbon dioxide and water. Does the ethanol undergo a physical change or a chemical c

NARA [144]

It's a chemical change because ethanol is converted into other substances.

3 0

2 years ago

Convert 30°C into kelvin

lakkis [162]

Answer:303.15

Explanation:

4 0

2 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.

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

Hence,

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

8 0

3 years ago

A student performed a chemical reaction in which 35 grams of hydrogen and 65 grams of oxygen reacted to form water. What is the

sergejj [24]

Answer:

The mass of the product is 73,06 g

Explanation:

The reaction to form water is:

2H2 + O2 ---> 2H2O

35 g 65 g

Molar mass H2 = 2g/m ---> I have 17,5 moles (35g / 2 g/m)

Molar mass O2 = 32g/m ---> I have 2,03 moles (65g / 32g/m)

2 moles of H2 __reacts with __ 1 mol O2

17,5 moles of H2 __ reacts with ___ (17,5 m . 1 m) / 2 m = 8,75 m

I have just 2,03 m of O2 and I need 8,75 m so O2 is my limiting reagent

1 mol of O2 ___ reacts with 2 moles H2

2,03 moles of O2 __ reacts with (2,03 m . 2 m) / 1 m = 4,06 m

I have 17,5 moles of H2 and I need 4,06 m. H2 is my excess reagent

"All operations are done with the limiting reagent"

1 mol O2 are required____ to form 2 H2O

2,03 mol O2 are required __ to form (2,03m . 2m) / 1m = 4,06 m

Molar mass of water: 18 g/m

Mass of water ----> Molar mass . Moles water = 18 g/m . 4,06 m = 73,08g

8 0

2 years ago

Some athletes have as little as 3.0% body fat. If such a person has a body mass of 67 kg , how many pounds of body fat does that

Elanso [62]

Data:
mass = 67 Kg → 147.71 pounds

Solving:

$\frac{3}{100}*147.71 = \frac{443.13}{100} \approx \boxed{4.43\:pounds\:of\:body\:fat}\end{array}}\qquad\quad\checkmark$

8 0

3 years ago