Answer:
B. Cementation
Explanation:
The processes by which sediments are changed into rock are complex, but can be simplified into two processes, called compaction and cementation. Rounded sediment grains (ooliths) bound together with crystalline calcite.
K= [Ar] 4s^1
Pb= [Xe] 6s^2 5d^10 6p^2
Sc= [Ar] 4s^2 3d^1
Ra= [Rn] 7s^2
O= [He] 2s^2 2p^4
Ag= [Kr] 5s^2 4d^9
Ru= [Kr] 5s^1
Ce= [Xe] 6s^2 5d^1 4f^1
I= [Kr] 5s^2 4d^10 5p^5
F= [He] 2s^2 2p^5
Answer:
The percentage composition of the elements of the compound in the three samples is the same.
Explanation:
<em>The law of definite proportions states that all pure samples of a particular chemical compound contain the same elements in the same proportion by mass.</em>
Sample A:
Mass of A = 4.31 g; mass of Z = 7.70 g
Total mass of sample = 12.01
Percentage mass of A in the sample = (4.31 * 100)/12.01 = 35.9 %
Percentage mass of Z in the sample = (7.70 * 100)/12.01 = 64.1 %
Sample B:
Percentage mass of A in the sample = 35.9 %
Percentage mass of Z in the sample = 64.1 %
Sample C:
Mass of A = 0.718 g; Total mass of sample = 2.00 g
mass of Z = mass of sample - mass of A = 2.00 g - 0.718 g = 1.282 g
Percentage mass of A in the sample = (0.718 * 100)/2.00 = 35.9 %
Percentage mass of Z in the sample = (1.282 * 100)/2.00 = 64.1 %
From the calculations, it can be seen that the percentage composition of the elements in the compound is the same for the three samples.
Answer:
Explanation:
Assume that you have mixed 135 mL of 0.0147 mol·L⁻¹ NiCl₂ with 190 mL of 0.250 mol·L⁻¹ NH₃.
1. Moles of Ni²⁺
2. Moles of NH₃
3. Initial concentrations after mixing
(a) Total volume
V = 135 mL + 190 mL = 325 mL
(b) [Ni²⁺]
(c) [NH₃]
3. Equilibrium concentration of Ni²⁺
The reaction will reach the same equilibrium whether it approaches from the right or left.
Assume the reaction goes to completion.
Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺
I/mol·L⁻¹: 6.106×10⁻³ 0.1462 0
C/mol·L⁻¹: -6.106×10⁻³ 0.1462-6×6.106×10⁻³ 0
E/mol·L⁻¹: 0 0.1095 6.106×10⁻³
Then we approach equilibrium from the right.
Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺
I/mol·L⁻¹: 0 0.1095 6.106×10⁻³
C/mol·L⁻¹: +x +6x -x
E/mol·L⁻¹: x 0.1095+6x 6.106×10⁻³-x
Kf is large, so x ≪ 6.106×10⁻³. Then
Answer:
Explanation:
Hello,
In this case, given that 1 inch equals 2.54 cm and 1 lb equals 453.6 g we apply the following conversion factor in order to compute the required density in lb/in³:
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