Answer:
Khud karo warna teacher ko bata don ga
Explanation:
Answer:
Explanation:
Density can be found by dividing the mass by the volume.
The mass of the water is 50.0 grams.
The volume of the water is 50.0 milliliters.
Substitute the values into the formula.
Divide.
The density of the water is 1.0 grams per milliliter. Also, remember that the density of pure water is always 1.0 g/mL or g/cm³
Answer: The given statement is true.
Explanation:
When we increase the amount of solvent which is water in this case then it means there will occur an increase in the molecules. Hence, there will be more number of collisions to take place with increase in number of molecules.
Therefore, more is the amount of interaction taking place between the molecules of a solution more will be its rate of hydrolysis.
Thus, we can conclude that the statement increasing the amount of water in which the sugar is dissolved will increase the frequency of collisions between the sucrose molecules and the water molecules resulting in an increase in the rate of hydrolysis, is true.
Answer:
0.0055 mol of N2O5 will remay after 7 min.
Explanation:
The reaction follows a first-order.
Let the concentration of N2O5 after 7 min be y
Rate = Ky = change in concentration of N2O5/time
K is rate constant = 6.82×10^-3 s^-1
Initial concentration of N2O5 = number of moles/volume = 2.1×10^-2/1.8 = 0.0117 M
Change in concentration = 0.0117 - y
Time = 7 min = 7×60 = 420 s
6.82×10^-3y = 0.0117 - y/420
0.0117 - y = 420×6.82×10^-3y
0.0117 - y = 2.8644y
0.0117 = 2.8644y + y
0.0117 = 3.8644y
y = 0.0117/3.8644 = 0.00303 M
Number of moles of N2O5 left = y × volume = 0.00303 × 1.8 = 0.0055 mol (to 2 significant digits)
Answer:
25.907°C
Explanation:
In Exercise 102, heat capacity of bomb calorimeter is 6.660 kJ/°C
The heat of combustion of benzoic acid is equivalent to the total heat energy released to the bomb calorimeter and water in the calorimeter.
Thus:
= heat of combustion of benzoic acid
= heat energy released to water
= heat energy released to the calorimeter
Therefore,
![-m_{combust}*H_{combust} = [m_{water}*c_{water} + C_{calori}]*(T_{f} - T_{i})](https://tex.z-dn.net/?f=-m_%7Bcombust%7D%2AH_%7Bcombust%7D%20%3D%20%5Bm_%7Bwater%7D%2Ac_%7Bwater%7D%20%2B%20C_%7Bcalori%7D%5D%2A%28T_%7Bf%7D%20-%20T_%7Bi%7D%29)
1.056*26.42 = [0.987*4.18 + 6.66](
- 23.32)
27.8995 = [4.12566+6.660]( - 23.32)
( - 23.32) = 27.8995/10.7857 = 2.587
= 23.32 + 2.587 = 25.907°C
