14. The following are a list of the four parts of John Dalton's atomic theory Indicate which of

ch4aika [34]

Use the molar mass of ammonia to change the grams to moles and then use mole-mole ratio

100. g NH3 (1 mol NH3/ 17.04 g) (3 mol H2/ 2 mol NH)= 8.80 moles H2

5 0

3 years ago

For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia

Fittoniya [83]

Answer : The value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

To calculate the $E^o_{(Cu^{2+}/Cu)}$ of the reaction, we use the equation:

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

$E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}$

Putting values in above equation, we get:

$1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)$

$E^o_{(Cu^{2+}/Cu)}=0.34V$

Hence, the value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

8 0

3 years ago

Which of the following is a characteristic of a scientific theory

Andru [333]

Answer:

number one

Explanation:

hope I helped

8 0

3 years ago

How many grams of water, H20 are needed if 88<br> grams of CO2 gas are produced?

joja [24]

Answer:

Explanation:

If one mole of carbon monoxide has a mass of 28.01 g and one mole of carbon dioxide has a mass of 44.01 g , it follows that the reaction produces 44.01 g of carbon dioxide for every 28.01 g of carbon monoxide.

3 0

2 years ago

450g of chromium(iii) sulfate reacts with excess potassium phosphate. How many grams of potassium sulfate will be produced? (ANS

Irina18 [472]

Answer:

600 g K₂SO₄

Explanation:

First write down the complete, balanced chemical equation for such question. In this case:

Cr₂(SO₄)₃ + 2K₃PO₄ → 3K₂SO₄ + 2CrPO₄

Next, calculate molar masses of required compounds mentioned in the question. In this case it is for Cr₂(SO₄)₃ and K₂SO₄.

Molar mass(MM) of Cr₂(SO₄)₃:

= 2*(MM of Cr) + 3*( MM of S) + 3*4*( MM of O)

= 2*(52) + 3*(32) + 12*(16)

= 104 + 96 + 192

= 392 g

Molar mass(MM) of K₂SO₄:

= 2*(MM of K) + 1*( MM of S) + 4*( MM of O)

= 2*(39) + 32 + 4*(16)

= 78 + 32 + 64

= 174 g

Here comes the concept of Limiting reagent:

The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it. The other reactants present with this are usually in excess and called excess reactants. If quantities of both the reactants are given, then one should apply unitary method and find out the limiting reagent out of the two. Then, determine the amount of product formed or percentage yield.

Also, 1 mole( 392 g) of Cr₂(SO₄)₃ gives 3 moles( 174*3 = 522 g) of K₂SO₄.

Using unitary method, if 392g of Cr₂(SO₄)₃ gives 522 g of K₂SO₄ , then 450 g of Cr₂(SO₄)₃ will give how much of K₂SO₄?

Yeild of K₂SO₄ : $\frac{522 * 450}{392}$

That is 599.3 g.

Since we have not considered molecular masses of individual atoms to 6 decimal places, this number can be approximated to 600g.

Therefore, 600g of K₂SO₄ is produced.

6 0

3 years ago