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Temka [501]
3 years ago
7

An alpha particle is equivalent to a _____ nucleus.

Chemistry
2 answers:
EleoNora [17]3 years ago
7 0

Answer: B. helium

Explanation:

Alpha decay: In this process, alpha particles is emitted when a heavier nuclei decays into lighter nuclei. The alpha particle released has a charge of +2 units.

General representation of an element is given as:_Z^A\textrm{X} where,

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

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

As the alpha particle has atomic number of 2 and mass number of 4, this is helium nuclei.

astra-53 [7]3 years ago
6 0
Alpha particle is equivalent to B. Helium atom (2 protons, 2 neutrons) 
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​29. A gas has a volume of 1.75 L at -23°C and 150.0 kPa.
arsen [322]

The answer for the following mention bellow.

  • <u><em>Therefore the final temperature of the gas is 260 k</em></u>

Explanation:

Given:

Initial pressure (P_{1}) = 150.0 kPa

Final pressure (P_{2}) = 210.0 kPa

Initial volume (V_{1}) = 1.75 L

Final volume (V_{2}) = 1.30 L

Initial temperature (T_{1}) = -23°C = 250 k

To find:

Final temperature (T_{2})

We know;

According to the ideal gas equation;

P × V = n × R ×T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas  constant

T represents the temperature of the gas

We know;

\frac{P*V}{T} = constant

\frac{P_{1} }{P_{2} } × \frac{V_{1} }{V_{2} } = \frac{T_{1} }{T_{2} }

Where;

(P_{1}) represents the initial pressure of the gas

(P_{2}) represents the final pressure of the gas

(V_{1}) represents the initial volume of the gas

(V_{2}) represents the final volume of the gas

(T_{1}) represents the initial temperature of the gas

(T_{2}) represents the final temperature of the gas

So;

\frac{150 * 1.75}{210 * 1.30} = \frac{260}{T_{2} }

(T_{2}) =260 k

<u><em>Therefore the final temperature of the gas is 260 k</em></u>

<u><em></em></u>

3 0
3 years ago
It took 55 days for a radioactivity of 1.75 x 1012 Bq to remain 0.135 Ci. What is the half-life of this radioactivity?
Mnenie [13.5K]

From the calculations, the half life of the material is 6.5 days.

<h3>What is radioactivity?</h3>

The term radioactivity has to do with the spontaneous disintegration of a specie.

Uisng the formula;

N=Noe^-kt

N= amount at time t = 0.135 Ci or 4.995 ×10^9 Bq

No = amount initially present =  1.75 x 10^12 Bq

k = rate constant = ?

t = time taken = 55 days

Hence;

4.995 ×10^9  = 1.75 x 10^12e^-55k

4.995 ×10^9/1.75 x 10^12 = e^-55k

2.85 * 10^-3 = e^-55k

ln2.85 * 10^-3 = -55k

k = ln2.85 * 10^-3/-55

k = 0.1066 day-1

Half life = 0.693/ 0.1066 day-1

= 6.5 days

Learn more about radioactivity:brainly.com/question/1770619

#SPJ1

4 0
1 year ago
what method can be used to seperate parts of liquid mixture wwhen the entire mixture can pass through a filter
QveST [7]
Distillation is the <span>method that can be used to seperate parts of liquid mixture when the entire mixture can pass through a filter.</span>
6 0
3 years ago
Apart from plants, what other group of organisms photosynthesises?<br>​
Licemer1 [7]

Answer:

Plants, algae, and a group of bacteria called cyanobacteria are the only organisms capable of performing photosynthesis

4 0
2 years ago
When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II)
lesya692 [45]

Answer: The mass of lead deposited on the cathode of the battery is 1.523 g.

Explanation:

Given: Current = 62.0 A

Time = 23.0 sec

Formula used to calculate charge is as follows.

Q = I \times t

where,

Q = charge

I = current

t = time

Substitute the values into above formula as follows.

Q = I \times t\\= 62.0 A \times 23.0 sec\\= 1426 C

It is known that 1 mole of a substance tends to deposit a charge of 96500 C. Therefore, number of moles obtained by 1426 C of charge is as follows.

Moles = \frac{1426 C}{96500 C/mol}\\= 0.0147 mol

The oxidation state of Pb in PbSO_{4} is 2. So, moles deposited by Pb is as follows.

Moles of Pb = \frac{0.0147}{2}\\= 0.00735 mol

It is known that molar mass of lead (Pb) is 207.2 g/mol. Now, mass of lead is calculated as follows.

No. of moles = \frac{mass}{molar mass}\\ 0.00735 = \frac{mass}{207.2 g/mol}\\mass = 1.523 g

Thus, we can conclude that the mass of lead deposited on the cathode of the battery is 1.523 g.

3 0
2 years ago
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