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Semmy [17]
3 years ago
9

10.0 grams of water are heated during the preparation of a cup of coffee 1.0x 103 j of the heat are added to the water. which is

initially at 20 c what is the final temperature of the coffee
Chemistry
1 answer:
nirvana33 [79]3 years ago
8 0

<u>Answer:</u> The final temperature of the coffee is 43.9°C

<u>Explanation:</u>

To calculate the final temperature, we use the equation:

q=mC(T_2-T_1)

where,

q = heat released = 1.0\times 10^3J=1000J

m = mass of water = 10.0 grams

C = specific heat capacity of water = 4.184 J/g°C

T_2 = final temperature = ?

T_1 = initial temperature = 20°C

Putting values in above equation, we get:

1000J=10.0g\times 4.184J/g^oC\times (T_2-20)\\\\T_2=43.9^oC

Hence, the final temperature of the coffee is 43.9°C

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The vapor pressure of water and the partial pressure of hydrogen contribute to the total pressure of 715 torr. What is the parti
Lostsunrise [7]

Answer:

0.91 atm is the partial pressure of just hydrogen gas.

Explanation:

Vapor pressure of water , p= 0.0313 atm

Partial pressure of hydrogen gas = p_{H_2}

Total pressure of the water vapors and hydrogen gas = P = 715 Torr

1 atm = 715 Torr

715 Torr=\frac{715}{760} atm=0.94 atm

According Dalton's law of partial pressure:

P=p+p_{H_2}

0.94 atm=0.0313 atm+p_{H_2}

p_{H_2}=0.94 atm - 0.0313 atm =0.9087 atm \approx 0.91 atm

0.91 atm is the partial pressure of just hydrogen gas.

8 0
3 years ago
Polonium- 210 , Po210 , decays to lead- 206 , Pb206 , by alpha emission according to the equation Po84210⟶Pb82206+He24 If the ha
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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  

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3 years ago
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When chemicals were not mixed they were at room temperature and when we mix them exothermic reaction took place and heat was released which raised the temperature of mixture.

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Melting point: -115.6°F (-82°C)

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