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Artemon [7]
2 years ago
10

The Î""G°′ of the reaction is −7.180 kJ·mol−1. Calculate the equilibrium constant for the reaction at 25 °C.

Chemistry
1 answer:
nasty-shy [4]2 years ago
5 0

Answer:

Explanation:The Î""G°′ of the reaction is −7.180 kJ·mol−1. Calculate the equilibrium constant for the reaction at 25 °

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A common fuel additive that is composed of C, H, and O enhanced the performance of gasoline began being phased out in 1999 becau
spayn [35]

Answer:

C₅ H₁₂ O

Explanation:

44 g of CO₂ contains 12 g of C

30.2 g of CO₂ will contain 12 x 30.2 / 44 = 8.236 g of C .

18 g of H₂O contains 2 g of hydrogen

14.8 g of H₂0 will contain 1.644 g of  H .

total compound = 12.1 out of which 8.236 g is C and 1.644 g is H , rest will be O

gram of O = 2.22

moles of C, O, H in the given compound =  8.236 / 12 , 2.22 / 16 , 1.644 / 1

= .6863 , .13875 , 1.644

ratio of their moles = 4.946 : 1 : 11.84

rounding off to digits

ratio = 5 : 1 : 12

empirical formula = C₅ H₁₂ O

6 0
3 years ago
You used a calorimeter in the heat transfer lab. Explain how the calorimeter works, and how to calculate the heat given off or a
Olenka [21]
A calorimeter contains reactants and a substance to absorb the heat absorbed. The initial temperature (before the reaction) of the heat absorbent is measured and then the final temperature (after the reaction) is also measured. The absorbent's specific heat capacity and mass are also known. Given all of this data, the equation:
Q = mcΔT 
To find the heat released.
7 0
3 years ago
Read 2 more answers
The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of
timama [110]

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = 1.83g/cm^3=1.83g/ml

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml

Now we have to calculate the molarity of solution.

Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

7 0
3 years ago
Standard temperature and pressure are _____.
Maurinko [17]
The standard temperature is 0c.
the standard pressure is 1atm.
Mark As Brainliest
4 0
3 years ago
A certain radioactive nuclide has a half life of 1.00 hour(s). Calculate the rate constant for this nuclide. s-1 Calculate the d
Karo-lina-s [1.5K]

Answer:

k= 1.925×10^-4 s^-1

1.2 ×10^20 atoms/s

Explanation:

From the information provided;

t1/2=Half life= 1.00 hour or 3600 seconds

Then;

t1/2= 0.693/k

Where k= rate constant

k= 0.693/t1/2 = 0.693/3600

k= 1.925×10^-4 s^-1

Since 1 mole of the nuclide contains 6.02×10^23 atoms

Rate of decay= rate constant × number of atoms

Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms

Rate of decay= 1.2 ×10^20 atoms/s

8 0
2 years ago
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