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Katen [24]
3 years ago
12

When a 0.4500 g sample of impure potassium chloride was dissolved in water and treated with an excess ofsilver nitrate, 0.8402 g

of silver chloride was precipitated. Calculate the percentage KCl in the original sample.
Chemistry
1 answer:
Svetllana [295]3 years ago
5 0

Answer:

97.78% KCl in the original sample

Explanation:

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Complete combustion of 2.60 g of a hydrocarbon produced 8.46 g of co2 and 2.60 g of h2o. what is the empirical formula for the h
zubka84 [21]
1 mole of carbon dioxide contains a mass of 44 g, out of which 12 g are carbon. 
Hence, in this case the mass of carbon in 8.46 g of CO2:
 (12/44) × 8.46 = 2.3073 g
 1 mole of water contains 18 g, out of which 2 g is hydrogen;
Therefore, 2.6 g of water contains;
 (2/18) × 2.6 = 0.2889 g of hydrogen.
Therefore, with the amount of carbon and hydrogen from the hydrocarbon we can calculate the empirical formula.
We first calculate the number of moles of each,
Carbon = 2.3073/12  = 0.1923 moles
Hydrogen = 0.2889/1 = 0.2889 moles
Then, we calculate the ratio of Carbon to hydrogen by dividing with the smallest number value;
             Carbon : Hydrogen
  0.1923/0.1923 : 0.2889/0.1923
                       1 :  1.5
                      (1 : 1.5) 2
                     = 2 : 3
Hence, the empirical formula of the hydrocarbon is C2H3
5 0
4 years ago
The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate c
crimeas [40]

Answer : The rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 785.0K = ?

Ea = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 785.0K

Now put all the given values in this formula, we get:

\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]

K_2=1.45\times 10^{-2}s^{-1}

Therefore, the rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

7 0
4 years ago
When glucose is fermented, ethanol is formed together with ………………….
levacccp [35]

Answer:

carbon dioxide and oxygen

3 0
3 years ago
If all other variables remain unchanged, what happens to the output force when the area of the input piston is doubled?
mars1129 [50]

Answer:

1250N

Explanation:

This question is based on pascal's Law.

So By Pascal's Law

 =  

therefore  =force on input piston =25N

                  = Force or weight on output person.

therefore after putting the values we get,

= (25x 1500)/30

    =1250N

3 0
3 years ago
An atom has atomic number of 34. how many protons and electrons does it have? what is its symbol? 34 protons and electrons. se l
AURORKA [14]
Se: Selenium

Protons: 34
Electrons: All atoms in the periodic table are neutral until changed otherwise
Neutrons: 44
Atomic Mass: 78.09

>Symbol attached<

4 0
3 years ago
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