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son4ous [18]
3 years ago
15

How to write a balanced nuclear equation on how uranium can disintegrate to Actium​

Chemistry
1 answer:
andrey2020 [161]3 years ago
5 0

Answer:

Balancing Nuclear Equations

To balance a nuclear equation, the mass number and atomic numbers of all particles on either side of the arrow must be equal.

Explanation:

follows:

6

3

Li

+

2

1

H

→

4

2

He

+

?

To balance the equation above for mass, charge, and mass number, the second nucleus on the right side must have atomic number 2 and mass number 4; it is therefore also helium-4. The complete equation therefore reads:

6

3

Li

+

2

1

H

→

4

2

He

+

4

2

He

Or, more simply:

6

3

Li

+

2

1

H

→

2

4

2

He

image

Lithium-6 plus deuterium gives two helium-4s.: The visual representation of the equation we used as an example.

Compact

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Enzymatic breakdown of which of the following compounds doesn't begin until it reaches the stomach?A) lipidsB) carbohydrates C)
kvv77 [185]

Answer is proteins (C)

7 0
3 years ago
A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture co
DaniilM [7]

Answer:

Explanation:

The air 9% mole% methane have an average molecular weight of:

9%×16,04g/mol + 91%×29g/mol = 27,8g/mol

And a flow of 700000g/h÷27,8g/mol = 25180 mol/h

In the reactor where methane solution and air are mixed:

In = Out

Air balance:

91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)

Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>

<em></em>

The air in the product gas is

95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>

43058 mol air×29g/mol <em>1249 kg air</em>

Percent of oxygen is: \frac{289kg}{1249 kg} =<em>0,231 kg O₂/ kg air</em>

<em></em>

I hope it helps!

4 0
3 years ago
Which best describes the importance of the microscope to the development of cell theory? Microscopes color different parts of ce
aliya0001 [1]
Cells are too small to see with the naked eye. 

It's pretty straight forward, use the cross-out method. 

1) Microscopes MAGNIFY images, they don't color the cells. In fact, scientists have to use these chemicals to "stain" or color the cells to see them more easily through microscopes. 

2) If the lenses of a microscope reduced the image of an organism to the size of a cell, you'd be seeing a very tiny human through your microscope, instead of actual cells. 

3) Microscopes don't "trap" anything. In fact, scientists use plates or slides under microscopes to contain what they're studying. 

8 0
3 years ago
Read 2 more answers
Which statement is true about air temperature and humidity
lapo4ka [179]

Complete Question:

Which statement is true about air temperature and humidity?

Group of answer choices.

a. the air temperature does not affect how much moisture the air can hold

b. hotter air holds less moisture than colder air.

c. hot air and cold air share the same amount of moist.

d. colder air holds less moisture than hotter air.

Answer:

d. colder air holds less moisture than hotter air

Explanation:

Weather can be defined as the atmospheric conditions of a particular area over a short period of time.

The elements of weather include precipitation, wind, temperature, atmospheric pressure, relative humidity, cloud, and wind speed.

Temperature can be defined as a measure of the degree of coldness or hotness of a physical object (body).

On the other hand, humidity refers to the concentration (amount) of water vapor that is present in the air. It is high when there's a lot of water vapor in the air and low when the level of water vapor is small.

Hence, the true statement about air temperature and humidity is that colder air holds less moisture than hotter air because as the air cools, its molecules move closer together while the molecules move farther apart as the air become hot.

Additionally, at constant humidity, relative humidity is inversely proportional to temperature i.e as the temperature decreases, relative humidity increases.

6 0
3 years ago
Heyy guys, so basically i need help with stoichiometric calculation I will give you 100 points just to answer all of these answe
jeka94

Answer:

3. The mass of ethanol required is approximately 0.522869 g

The mass of ethanoic acid required is approximately 0.68156 g

4. The mass of iron (III) oxide required is approximately 285.952.189.095 tonnes

5. The mass of silver nitrate required is approximately 14.53 grams

6. The mass of copper oxide that would be needed is approximately 31.86 grams

7. a. The mass of the precipitate, Zn(OH)₂ formed is approximately 49.712 grams

b. The mass of the precipitate, Al(OH)₃ formed is approximately 13 grams

c. The mass of the precipitate, Mg(OH)₂, formed is approximately 14.579925 grams

Explanation:

3. The 1 mole of ethanol and 1 mole of ethanoic acid combines to form 1 mole of ethyl ethanoate

The number of moles of ethyl ethanoate in 1 gram of ethyl ethanoate, n = 1 g/(88.11 g/mol) = 1/88.11 moles

∴ The number of moles of ethanol = 1/88.11 moles

The number of moles of ethanoic acid = 1/88.11 moles

The mass of ethanol = (46.07 g/mol) × 1/88.11 moles = 0.522869 g

The mass of ethanoic acid in the reaction = 60.052 g/mol × 1/88.11 moles ≈ 0.68156 g

4. 1 mole of iron(III) oxide reacts with 1 mole of CO₂ to produce 1 mole of iron

The number of moles in 100 tonnes of iron= 100000000/55.845 = 1790670.60614 moles

The mass of iron (III) oxide required = 159.69 × 1790670.60614 = 285952189.095 g ≈ 285.952.189.095 tonnes

5. The number of moles of NaCl in 5 grams of NaCl = 5 g/58.44 g/mol = 0.0855578371 moles

The mass of silver nitrate required, m = 169.87 g/mol × 0.0855578371 moles ≈ 14.53 grams

6. The number of moles of CuSO₄·5H₂O in 100 g of CuSO₄·5H₂O = 100 g/(249.69 g/mol) ≈ 0.4005 moles

The mass of copper oxide required, m = 79.545 g/mol × 0.4005 moles ≈ 31.86 grams

7. a. The number of moles of NaOH in the reaction = 20 g/(39.997 g/mol) ≈ 0.5 moles

2 moles of NaOH produces 1 mole of Zn(OH)₂

0.5 moles of NaOH will produce 0.5 mole of Zn(OH)₂

The mass of 0.5 mole of Zn(OH)₂ = 0.5 mole × 99.424 g/mol = 49.712 grams

The mass of the precipitate, Zn(OH)₂ formed = 49.712 grams

b. 6 moles of NaOH produces 2 moles Al(OH)₃

20 g, or 0.5 mole of NaOH will produce (1/6) mole of Al(OH)₃

The mass of the precipitate, Al(OH)₃ formed, m = 78 g/mol×(1/6) moles = 13 grams

c. 2 moles of NaOH produces 1 mole of Mg(OH)₂, therefore;

20 g or 0.5 moles of NaOH formed (1/4) mole of Mg(OH)₂

The mass of the precipitate, Mg(OH)₂, formed, m = 58.3197 g/mol × (1/4) moles = 14.579925 grams

3 0
3 years ago
Read 2 more answers
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