23.0 + 60.0 = 83.0° C heat energy is required to raise
Answer:
0.15g
Explanation:
Given parameters:
Number of molecules of water = 1.2 x 10²¹ molecules
Unknown:
Mass of SnO₂ = ?
Solution:
To solve this problem, we have to work from the known to the unknown specie;
SnO₂ + 2H₂ → Sn + 2H₂O
Ensure that the equation given is balanced;
Now,
the known species is water;
6.02 x 10²³ molecules of water = 1 mole
1.2 x 10²¹ molecules of water = 
= 0.2 x 10⁻²moles
Number of moles of water = 0.002moles
From the balanced chemical equation:
2 mole of water is produced from 1 mole of SnO₂
0.002 moles of water will be produced from 
= 0.001moles
To find the mass;
Mass = number of moles x molar mass
Molar mass of SnO₂ = 118.7 + 2(16) = 150.7g/mol
Mass = 0.001 x 150.7 = 0.15g
Answer:
Explanation:
M=D times V
Answer-3,633.84g
Rounded Answer (correct sig figs)- 3600g
Answer:
electron-electron repulsion
Explanation:
When electrons add into valence shell of neutral elements, the element assumes a negative oxidation state. With this, the number of electrons having (-) charges will be larger than the number of protons having positive (+) charges. As a result, the extra electrons repel one another (i.e., like charges repel) and a larger radius is the result.
In contrast, when cations are formed, electrons are removed from the valence level (oxidation) producing an element having a greater number of protons than electrons. The larger number of protons will function to attract the electron cloud with a greater force that results in a contraction of atomic radius and a smaller spherical volume than the neutral unionized element.
To visualize, see attached chart that shows atomic and ionic radii before and after ionization of the elements.
The answer is potassium. It would be 4, and for neon would be 2. Just total which row of the periodic table you are on. The "L" tells you whether the highest-energy electron is in an "s" orbital (L=0) or a "p" orbital (L=1) or a "d" orbital (L=2) or an "f" orbital (L=3). The way in which these orbitals are filled is: for each of the first three rows (up to argon), two electrons in the "s" orbital are filled first, then 6 electrons in the "p"orbitals. The row where the potassium also starts with filling the "s" orbital at the new "n" level (4) but then goes back to satisfying up the "d" orbitals of n=3 before it seals up the "p"s for n=4.
