Answer:
Liquid A is a homogeneous mixture and Liquid B is a heterogeneous mixture
Explanation:
Answer:
1s22s22p6: Neon (Ne)
1s22s22p63s23p3: Phosphorous (P)
1s22s22p63s23p64s1: Potassium (K)
1s22s22p63s23p64s2(im not sure what 308 is supposed to be): Calcium (Ca)
1s22s22p63s23p64s23d104p65s24d3: there is no pure element that ends 4d3 that I know of so this can either be Zirconium(Zr) if it ends in 4d2 or Niobium (Nb) if it ends in 4d4
Explanation:
you can look at the periodic table and the trends to find the rough idea of where the electron configuration ends, there are helpful articles and images on these, i attached an image that may help. After that you can look at the atomic number to find the number of electrons for a pure element and use the electron subshell pattern thing to find the exact number
Answer:
Albert Einstein is perhaps most famous for introducing the world to the equation E=mc2. In essence, he discovered that energy and mass are interchangeable, setting the stage for nuclear power—and atomic weapons. His part in the drama of nuclear war may have ended there if not for a simple refrigerator.
Explanation:
Albert Einstein is perhaps most famous for introducing the world to the equation E=mc2. In essence, he discovered that energy and mass are interchangeable, setting the stage for nuclear power—and atomic weapons. His part in the drama of nuclear war may have ended there if not for a simple refrigerator.
Given:
Iron, 125 grams
T
1 = 23.5 degrees Celsius, T2 =
78 degrees Celsius.
Required:
Heat produced in kilojoules
Solution:
The molar mass of iron is 55.8
grams per mole. SO we need to change the given mass of iron into moles.
Number of moles of iron = 125 g/(55.8
g/mol) = 2.24 moles
<span>
Q (heat) = nRT = nR(T2 = T1)</span>
Q (heat) = 2.24 moles (8.314
Joules per mol degrees Celsius) (78.0 degrees Celsius – 23.5 degrees Celsius)
<u>Q (heat) = 1014.97 Joules or
1.015 kilojoules</u>
<span>This is the amount of heat
produced in warming 125 g f iron.</span>
Answer:

Explanation:
Hello,
In this case, since at 60 °C, 108 grams of ammonium bromide are completely dissolved in 100 grams of water for a saturated solution, once it is cooled to 30 °C, wherein only 83.2 grams are completely dissolved in 100 grams of water, the following mass will precipitate:

Best regards.