Answer:
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Explanation:
<u>Step 1: </u>Data given
mass of water = 300 grams
initial temperature = 10°C
final temperature = 50°C
Temperature rise = 50 °C - 10 °C = 40 °C
Specific heat capacity of water = 4.184 J/g °C
<u>Step 2:</u> Calculate the heat
Q = m*c*ΔT
Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)
Q = 50208 Joule = 50.2 kJ
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Hello!
The mass number in isotope notation is denoted A, the atomic number is denoted as Z, and the element is denoted as X.
In the given isotope, the mass of the isotope is 212 amu, and the atomic number is 82.
We know that the number of electrons, and protons are equal to the atomic number. Therefore, there are 82 protons. Also, to find the number of neutrons, we subtract the atomic number from the atomic mass.
212 - 82 = 130 neutrons
<u>Final answers</u>:
- Atomic Number: 82
- Mass number: 212
- Number of Protons: 82
- Number of Neutrons: 130
Answer:
A. The average of all the data points
Answer:
0.3758moles
Explanation:
moles of kcl = mass of kcl/ molar mass of kcl = 28/74.5 = 0.3758moles
212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
Option A.
Explanation:
Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.
So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.
Then, 
So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
