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kotegsom [21]
3 years ago
8

Ano ang kaganapan na nangyari at nasabing nagsimula ang holy ramon empire​

Chemistry
1 answer:
Ksenya-84 [330]3 years ago
5 0

Answer:

plzz give the question in English

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A gas at constant volume has a pressure of 4. 50 atm at 200. K. What will be the pressure of the gas at 250. K? 3. 60 atm 4. 60
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5.625 atm will be the pressure of the gas at 250 K temperature of the gas at constant volume.

<h3>How we calculate the pressure of the gas?</h3>

Pressure of the gas will be calculated by using the ideal gas equation as:

PV = nRT,

From the question, it is clear that:

Moles of the gas and volume is constant here, so we calculate the pressure by rearranging the above equation as:

P/T = nR/V

And required equation will be:

P₁/T₁ = P₂/T₂, where

P₁ = pressure of gas = 4.50 atm

T₁ = temperature of gas = 200 K

P₂ = pressure of gas = to find?

T₂ = temperature of gas = 250 K

On putting all these values in the above equation, we get

P₂ = 4.50 × 250 / 200 = 5.625 atm

Hence, 5.625 atm is the pressure of the gas.

To know more about ideal gas equation, visit the below link:

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A chemical or physical change that takes in heat from its surroundings is an exothermic process an explosion an endothermic proc
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Is this a true or false question?
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Both new moon and lunar eclipse occur when the moon is in the same plane the sun and the earth. However, both are different because of the relative position of the moon in its orbit – relative to earth. In a new moon, the moon is in between the earth and the sun, while in a lunar eclipse, the earth us in between the moon and the sun.

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Polonium- 210 , Po210 , decays to lead- 206 , Pb206 , by alpha emission according to the equation Po84210⟶Pb82206+He24 If the ha
elixir [45]

Answer:

0.269 g

Explanation:

Po_{210} ^{84}  ⟶  Pb_{206} ^{82}+  + He_{4} ^{2}

Data:

Half-life of  Po_{210} ^{84}  (T(1/2)) = 138.4 days

Mass of PoCl4 = 561 mg (0,561 g) and molecular weight of PoCl4 = 350. 79 g/mol  

Time = 338.8 days  

Isotopic masses  

Po_{210} ^{84} = 209.98 g/mol  

Pb_{206} ^{82} = 205.97 g/mol  

Concepts  

Avogadro’s number: This is the number of constituent particles that are contained in a mol of any substance. These constituted particles can be atoms, molecules or ions). Its value is 6.023*10^23.  

The radioactive decay law is  

N=Noe^(-λt)

Where:  

No = number of atoms in t=0

N = number of atoms in t=t (now) in this case t=338.8 days  

λ= radioactive decay constant  

The radioactive constant is related to the half-life by the next equation  

λ= \frac{ln 2}{t(1/2)}

so  

λ= \frac{ln2}{138.4 days}  =0,005008 days^(-1)

No (Atoms of  Po_{210} ^{84}  in t=0)

To get No we need to calculate the number of atoms of  Po_{210} ^{84}   in the initial sample. We have a sample of 0,561 g of PoCl4. If we get the number of moles of PoCl4 in the sample, this will be the number of moles of  Po_{210} ^{84}  in the initial sample.  

This is:

\frac{0,561 g of PoCl4}{350. 79 g of PoCl4 /mol} = 0,001599 mol of  PoCl4

This is the number of mol of  Po_{210} ^{84} in the initial sample.

To get the number of atoms in the initial sample we use the Avogadro’s number = 6.023*10^23  

0,001599 mol of  Po_{210} ^{84} * 6.023*10^23 atoms/ mol of  Po_{210} ^{84} = 9.632 *10^20 atoms of  Po_{210} ^{84}

Atoms after 338.8 days

We use the radioactive decay law to get this value  

N=Noe^(-λt)

N=9.632*10^20 e^(-0,005008 days^(-1) * 338.8 days) =1.765*10^20

This is the number of atoms of  Po_{210} ^{84} in the sample after 338.8 days has passed  

The number of atoms  Po_{210} ^{84} transformed is equal to the number of atoms of Pb_{206} ^{82}  produced.  

The number of atoms of Po_{210} ^{84} transformed is No - N  

9.632 *10^20 – 1.765 *10^20 = 7.866*10^20

So, 7.866*10^20 is the number of atoms of Pb_{206} ^{82} produced  

We can get the mass with the Avogadro’s number

(7.866*10^20 atoms of Pb_{206} ^{82} ) / ( 6.023*10^23 atoms of Pb_{206} ^{82} / mol of Pb_{206} ^{82} =  0.001306 moles of Pb_{206} ^{82}

This number of moles have a mass of:

(0,001306 moles of Pb_{206} ^{82} )* (205.97 g of Pb_{206} ^{82} /mol of Pb_{206} ^{82} ) = 0.269 g  

3 0
3 years ago
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