<h3>
1.</h3>
C) The volume of the gas is proportional to the number of moles of gas particles.
The Avogadro's law applies to ideal gases with constant pressure and temperature. By that law, the volume of an ideal gas is proportional to the number of moles of particles in that gas.
<h3>2.</h3>
B) The gas now occupies less volume, and the piston will move downward.
Boyle's Law applies to ideal gases with a constant temperature. The volume of an ideal gas is inversely related to its pressure. A high pressure drives gas particles together, such that they occupy less volume. The gas trapped inside the piston has a smaller volume. As a result, the the piston will move downward.
Alternatively, consider the forces acting on the piston. Both the atmosphere and gravity are dragging the piston down. In order for it to stay in place, the gas below it must exert a pressure to balance the two forces. Now the pressure from outside has increased. The gas inside needs to increase its pressure. It needs a smaller volume to create that extra pressure. As a result, its volume will decrease, and the piston will move downwards.
<h3>
Answer:</h3>
0.0157 g Au
<h3>
General Formulas and Concepts:</h3>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.113 g Au
<u>Step 2: Identify Conversions</u>
Molar Mass of Au - 197.87 g/mol
<u>Step 3: Convert</u>
<u />
= 0.015733 g Au
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
0.015733 g Au ≈ 0.0157 g Au
A part of organism that is typically self contained and have a specific vital
Missing in your question :
Ksp of(CaCO3)= 4.5 x 10 -9
Ka1 for (H2CO3) = 4.7 x 10^-7
Ka2 for (H2CO3) = 5.6 x 10 ^-11
1) equation 1 for Ksp = 4.5 x 10^-9
CaCO3(s)→ Ca +2(aq) + CO3-2(aq)
2) equation 2 for Ka1 = 4.7 x 10^-7
H2CO3 + H2O → HCO3- + H3O+
3) equation 3 for Ka2 = 5.6 x 10^-11
HCO3-(aq) + H2O(l) → CO3-2 (aq) + H3O+(aq)
so, form equation 1& 2&3 we can get the overall equation:
CaCO3(s) + H+(aq) → Ca2+(aq) + HCO3-(aq)
note: you could get the overall equation by adding equation 1 to the inverse of equation 3 as the following:
when the inverse of equation 3 is :
CO3-2 (aq) + H3O+ (aq) ↔ HCO3- (aq) + H2O(l) Ka2^-1 = 1.79 x 10^10
when we add it to equation 1
CaCO3(s) ↔ Ca2+(aq) + CO3-2(aq) Ksp = 4.5 x 10^-9
∴ the overall equation will be as we have mentioned before:
when H3O+ = H+
CaCO3(s) + H+(aq) ↔ Ca2+ (aq) + HCO3-(aq) K= 80.55
from the overall equation:
∴K = [Ca2+][HCO3-] / [H+]
when we have [Ca2+] = [HCO3-] so we can assume both = X
∴K = X^2 / [H+]
when we have the PH = 5.6 so we can get [H+]
PH = - ㏒[H+]
5.6 = -㏒[H]
∴[H] = 2.5 x 10^-6
so, by substitution on K expression:
∴ 80.55 = X^2 / (2.5 x10^-6)
∴X = 0.0142
∴[Ca2+] = X = 0.0142
Hen two electrons occupy the same orbital, they always have opposite spin. No other knowledge about electron spin is needed for A-level.) Only when the 2s is filled, can the 2p sub-level fill. With the next six elements (boron, carbon, nitrogen, oxygen, fluorine and neon) the three 2p orbitals are filled.
