A gram of gasoline produces 45.0 kJ of energy when burned. Gasoline has a density of 0.77 g/ml. How would you ca the amount of e

Answer:- 64015 J

Solution: The calorimeter contains 4250 mL of water, which is at a temperature of 22.55 degrees Celsius.

The water's density is 1 gram per mL.

Thus, the mass of water = = 4250 grams.

After introducing the hot copper bar, the final temperature of the water reaches 26.15 degrees Celsius.

Thus, for the water = 26.15 - 22.55 = 3.60 degrees Celsius.

The specific heat capacity of water is 4.184 .

To determine the heat absorbed by the water, we can use the following formula:

where q represents heat energy, m refers to mass, and c indicates specific heat.

Now let's substitute the values into the equation to perform the calculations:

q = 64015 J

Therefore, the water absorbs 64015 J of heat.

Answer:

Explanation:

a) Iron has the tendency to undergo rusting -- this is a chemical property as it involves a reaction with water and air.

b) Precipitation in industrialized areas often has an acidic nature -- this is also a chemical property due to its interaction with bases or metals.

c) Hemoglobin is red in color -- this is a physical property since it doesn't entail any chemical reactions.

d) When water is left out in sunlight, it evaporates gradually -- this is a physical property because the process can easily be reversed, classifying it as a physical change.

e) During photosynthesis, plants convert carbon dioxide into more complex molecules -- this demonstrates a chemical property since it involves chemical reactions.

-0° is the freezing temperature of ice or ice cream.

The molar masses for H2S and NH3 are 34 and 17 g/mol, respectively. The appropriate equation to represent this is,
                                  Rate A/Rate B = √(molar mass B/molar mass A)
Substituting values,
                               x/77 = √(17 /34 )
                                 x = 54.4
This means NH3 will take 54.4 seconds to escape through the container. 

Answer:

7.3

Explanation:

Using the Henderson Hasselbalch equation, one can determine the pH or pOH of a solution via its pKa. Remember, , and pKa = -logKa, where Ka denotes the acid's equilibrium constant.

The Henderson Hasselbalch formula:

In this context, acid X possesses two ionic forms: the carboxyl group and an alternative form. Initially, we have 0.1 mol/L of acid in 100 mL, which gives:

n1 = (0.1 mol/L)×(0.1 L) = 0.01 mol

Upon dissociation, it yields 0.005 mol of the carboxyl form and 0.005 mol of the other form with stoichiometry assumed constant.

Introducing NaOH at a concentration of 0.1 mol/L and 75 mL, the moles of become:

n2 = (0.1 mol/L)×(0.075 L) = 0.0075 mol

Thus, 0.0075 mol of reacts with 0.005 mol of the carboxyl form, leading to 0.0025 mol of , which in turn reacts with 0.005 mol of the alternating group, leaving 0.0025 mol of the latter.

The new solution’s volume is 175 mL, but the concentrations of both forms remain unchanged in volume, so we can utilize the moles in the equation.

<pNote, the moles of the acid form remain 0.01 mol as it doesn’t undergo reaction!

Thus, we arrive at:

6.72 = pKa - log 4

pKa - log 4 = 6.72

pKa = 6.72 + log 4

pKa = 6.72 + 0.6

pKa = 7.3