c the temperature is increased?5)
Answer:
1) 0.0025 mol/L.s.
2) 0.0025 mol/L.s.
Explanation:
<em>H₂ + Cl₂ → 2HCl.</em>
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<em>The average reaction rate = - Δ[H₂]/Δt = - Δ[Cl₂]/Δt = 1/2 Δ[HCl]/Δt</em>
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<em>1. Calculate the average reaction rate expressed in moles H₂ consumed per liter per second.</em>
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The average reaction rate expressed in moles H₂ consumed per liter per second = - Δ[H₂]/Δt = - (0.02 M - 0.03 M)/(4.0 s) = 0.0025 mol/L.s.
<em>2. Calculate the average reaction rate expressed in moles CI₂ consumed per liter per second.</em>
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The average reaction rate expressed in moles Cl₂ consumed per liter per second = - Δ[Cl₂]/Δt = - (0.04 M - 0.05 M)/(4.0 s) = 0.0025 mol/L.s.
Answers:
(a) 1s² 2s²2p³; (b) 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d²; (c) 1s² 2s²2p⁶ 3s²3p⁵
Step-by-step explanation:
One way to solve this problem is to add electrons to the orbitals one-by-one until you have added the required amount.
Fill the subshells in the order listed in the diagram below. Remember that an s subshell can hold two electrons, while a p subshell can hold six, and a d subshell can hold ten.
(a) <em>Seven electrons
</em>
1s² 2s²2p³
There are two electrons in the 2s subshell and three in the 2p subshell. The remaining two electrons are in the inner 1s subshell.
(b) <em>22 electrons
</em>
1s² 2s²2p⁶ 3s²3p⁶ 4s²3d²
There are two electrons in the 4s subshell and two in the 2p subshell. The remaining 18 electrons are in the inner subshells.
(c) <em>17 electrons</em>
1s² 2s²2p⁶ 3s²3p⁵
There are two electrons in the 3s subshell and five in the 2p subshell. The remaining 10 electrons are in the inner subshells.
Answer:
b. have the same kind number of complete shells
Answer:
4.43 g of Oxygen
Explanation:
As shown in Chemical Formula, one mole of Aluminium Sulfate [Al₂(SO₄)₃] contains;
2 Moles of Aluminium
3 Moles of Sulfur
12 Moles of Oxygen
Also, the Molar Mass of Aluminium Sulfate is 342.15 g/mol. It means,
342.15 g ( 1 mole) of Al₂(SO₄)₃ contains = 192 g (12 mole) of O
So,
7.9 g of Al₂(SO₄)₃ will contain = X g of O
Solving for X,
X = (7.9 g × 192 g) ÷ 342.15 g
X = 4.43 g of Oxygen
