Answer:
1.atomic number
2.electron
3.element
4.atom
5.neutron
6.nucleus
7.proton
Explanation:
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Answer:
Moles of silver iodide produced = 1.4 mol
Explanation:
Given data:
Mass of calcium iodide = 205 g
Moles of silver iodide produced = ?
Solution:
Chemical equation:
CaI₂ + 2AgNO₃ → 2AgI + Ca(NO₃)₂
Number of moles calcium iodide:
Number of moles = mass/ molar mass
Number of moles = 205 g/ 293.887 g/mol
Number of moles = 0.7 mol
Now we will compare the moles of calcium iodide with silver iodide.
CaI₂ : AgI
1 : 2
0.7 : 2×0.7 = 1.4
Thus 1.4 moles of silver iodide will be formed from 205 g of calcium iodide.
Which of the following measurements is expressed to three significant figures?
C. 5.60 km
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In an ideal gas, there are no attractive forces between the gas molecules, and there is no rotation or vibration within the molecules. The kinetic energy of the translational motion of an ideal gas depends on its temperature. The formula for the kinetic energy of a gas defines the average kinetic energy per molecule. The kinetic energy is measured in Joules (J), and the temperature is measured in Kelvin (K).
K = average kinetic energy per molecule of gas (J)
kB = Boltzmann's constant ()
T = temperature (k)
Kinetic Energy of Gas Formula Questions:
1) Standard Temperature is defined to be . What is the average translational kinetic energy of a single molecule of an ideal gas at Standard Temperature?
Answer: The average translational kinetic energy of a molecule of an ideal gas can be found using the formula:
The average translational kinetic energy of a single molecule of an ideal gas is (Joules).
2) One mole (mol) of any substance consists of molecules (Avogadro's number). What is the translational kinetic energy of of an ideal gas at ?
Answer: The translational kinetic energy of of an ideal gas can be found by multiplying the formula for the average translational kinetic energy by the number of molecules in the sample. The number of molecules is times Avogadro's number:
Answer:
a) Kb = 10^-9
b) pH = 3.02
Explanation:
a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:
[NaA] and [A-] = 0.05/0.6 = 0.083 M
Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9
b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:
[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M
pKb = 10^-9
Ka = 10^-5
HA = H+ + A-
Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])
[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0
Clearing [H+]:
[H+] = 0.00095 M
pH = -log([H+]) = -log(0.00095) = 3.02
