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Serjik [45]
3 years ago
13

Which form of emission is commonly not written in nuclear equations because they do not affect charges, atomic numbers, or mass

numbers?
a. alpha particle
b. beta particle
c. gamma ray
Chemistry
2 answers:
lora16 [44]3 years ago
8 0
<span>The form of emission that is commonly not written in nuclear equations because they do not affect charges, atomic numbers, or mass numbers is the gamma ray. 

Gamma rays have no electrical charge nor atomic mass. It is only used to balance in a nuclear reaction, but gamma ray is considered powerful.

So, letter C. is the answer.
Hope this helps!</span>
schepotkina [342]3 years ago
4 0

Answer: c. gamma ray

Explanation: There are various process in which a radioactive nuclei decays:

1) Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units.

_A^Z\textrm{X}\rightarrow _{A-2}^{Z-4}+_2^4\alpha

A= mass number

Z= atomic number

2) Beta particle: It is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. This is also known as \beta ^+-decay. In this the mass number remains same.

_A^Z\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0e

3) Gamma ray emission: It is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. This decay releases \gamma -radiations. This process does not change the mass number.

_A^Z\textrm{X}^*\rightarrow _A^Z\textrm{X}+_0^0\gamma

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Consider the reaction of solid aluminum iodide and potassium metal to form solid potassium iodide and aluminum metal.The balance
ratelena [41]

Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

Percentage yield of Al = 67.8%

Actual yield of Al = 30.25 g

Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

67.8% = 30.25 / Theoretical yield

67.8 / 100 = 30.25 / Theoretical yield

0.678 = 30.25 / Theoretical yield

Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

Theoretical yield = 30.25 / 0.678

Theoretical yield of Al = 44.62 g

Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

AlI₃(s) + 3K(s) → 3KI(s) + Al(s)

Molar mass of AlI₃ = 27 + (3×127)

= 27 + 381 = 408 g/mol

Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 1 × 27 = 27 g

Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

8 0
3 years ago
After he finished his work, Mendeleev predicted _____.
Alex17521 [72]

Answer:

the properties of three undiscovered elements  is the correct answer.

Explanation:

After he completed his work Mendeleev predicted the properties of the undiscovered elements.

In the year 1871, Dmitri Mendeleev predicted the detailed existence and the properties of the three elements that are undiscovered and he gained tremendous fame.

He left the gap in the table to put the elements which are not identified at that time and by glancing at the physical and chemical properties of an element next to the gap, Mendeleev could predict the properties of the not discovered elements.

4 0
3 years ago
I need help with 18-23
Bess [88]

Answer:

18. Liters

19. grams

20. meters

21. meters

22. liters or mass (not sure on this one) it depends

23. grams

7 0
3 years ago
You have a different rock with a volume of 30cm and a mass of 60g, What is its density?
Galina-37 [17]
Density = mass/volume

Therefore,
Density = 60g/30cm
8 0
3 years ago
The concentration of hydrogen peroxide solution can be determined by
max2010maxim [7]

The question is incomplete, the complete reaction equation is;

The concentration of a hydrogen peroxide solution can be determined by titration

with acidified potassium manganate(VII) solution. In this reaction the hydrogen

peroxide is oxidised to oxygen gas.

A 5.00 cm3 sample of the hydrogen peroxide solution was added to a volumetric flask

and made up to 250 cm3 of aqueous solution. A 25.0 cm3 sample of this diluted

solution was acidified and reacted completely with 24.35 cm3 of 0.0187 mol dm–3

potassium manganate(VII) solution.

Write an equation for the reaction between acidified potassium manganate(VII)

solution and hydrogen peroxide.

Use this equation and the results given to calculate a value for the concentration,

in mol dm–3, of the original hydrogen peroxide solution.

(If you have been unable to write an equation for this reaction you may assume that

3 mol of KMnO4 react with 7 mol of H2O2. This is not the correct reacting ratio.)

Answer:

2.275 M

Explanation:

The equation of the reaction is;

2 MnO4^-(aq) + 16 H^+(aq) + 5H2O2(aq) -------> 2Mn^+(aq) + 10H^+ (aq) + 8H2O(l)

Let;

CA= concentration of MnO4^- =  0.0187 mol dm–3

CB = concentration of H2O2 = ?

VA = volume of MnO4^- = 24.35 cm3

VB = volume of H2O2 = 25.0 cm3

NA = number of moles of  MnO4^- = 2

NB = number of moles of H2O2 = 5

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB = CAVANB/VBNA

CB = 0.0187 * 24.35 * 5/25.0 * 2

CB = 0.0455 M

Since  

C1V1 = C2V2

C1 = initial concentration of H2O2 solution = ?

V1 = initial volume of H2O2 solution =  5.0 cm3

C2 = final concentration of H2O2 solution= 0.0455 M

V2 = final volume of H2O2 solution = 250 cm3

C1 = C2V2/V1

C1 = 0.0455 * 250/5

C1 = 2.275 M

8 0
3 years ago
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