Answer:
See explanation below
Explanation:
To solve this problem, we need to use the expression of half life decay of concentration (or mass) which is the following:
m = m₀e^-kt (1)
In this case, k will be the constant rate of this element. This is calculated using the following expression:
k = ln2/t₁/₂ (2)
Let's calculate the value of k first:
k = ln2/2.7 = 0.2567 d⁻¹
Now, we can use the expression (1) to calculate the remaining mass:
m = 8.1 * e^(-0.2567 * 2.6)
m = 8.1 * e^(-0.6674)
m = 8.1 * 0.51303
m = 4.16 mg remaining
Answer:
According to Hund's rule and the Aufbau principle in which the orbitals must be filled with electrons, they are not strictly applied in the real universe, because the intermediate and electron-filled atomic orbitals are very stable . Because there are four d-orbitals in universe L, a typical half-full configuration will be xd4 and its full configuration will be xd8, where x is the primary orbital for any specific element. Here is an example:
Vahadium ₂₃V
in real universe: [Ar]₈ 3d³4s²
in universe L: [Ar]₁₈ 3d⁴4s¹
Chromium
in real universe: [Ar]₈ 3d⁵4s¹
in universe L: [Ar]₁₈ 3d⁴4s²
Explanation:
To the right
Because there is an unbalanced force in that direction
:)
Answer:
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Answer:
1461.7 g of AgI
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI
Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:
Mole of AgI = 6.22 moles
Molar mass of AgI = 108 + 127
= 235 g/mol
Mass of AgI =?
Mass = mole × molar mass
Mass of AgI = 6.22 × 235
Mass of AgI = 1461.7 g
Therefore, 1461.7 g of AgI were obtained from the reaction.
