there is a system like evaporation the water water evaporates to give condensation is were all the water forms into a cloud and when it gets a lot of water you get persipition is when it rains
Answer:
(a) ₁₉K: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
(b) ₁₀Ne: 1s² 2s² 2p⁶
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(a) 3
(b) 6
(c) 7
Explanation:
We can state the ground-state electron configuration for each element following Aufbau's principle.
(a) ₁₉K: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
(b) ₁₀Ne: 1s² 2s² 2p⁶
Second part
(a) Al belongs to Group 13 in the Periodic Table. It has 13-10=3 electrons in the valence shell.
(b) O belongs to Group 16 in the Periodic Table. It has 16-10=6 electrons in the valence shell.
(c) F belongs to Group 17 in the Periodic Table. It has 17-10=7 electrons in the valence shell.
Answer:
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Explanation:
Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.
H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
We will balance it using the trial and error method.
First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
Finally, we will get the balanced equation by multiplying H₂O by 6.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Answer:
Manganese (II) ion, Mn²⁺
Explanation:
Hello,
In this case, given the overall reaction:
Thus, since manganese (II) ion, Mn²⁺ is both at the reactant and products, we infer it is catalyst, since catalysts are firstly consumed but finally regenerated once the reaction has gone to completion. Moreover, since inner steps are needed to obtain it, we can infer that the given rate law corresponds to the slowest step that is related with the initial collisions between Ce⁴⁺ and Mn²⁺
Best regards.
Answer:
Mass of glucose = 515.34 g
Explanation:
We are given;
Mass; m = 60 kg
Elevation; h = 1550 m
Acceleration due to gravity; 9.8 m/s²
Now, work performed to lift 60kg by 1550m is given by the formula;
W = mgh
W = 60 × 9.8 × 1550
W = 911400 J
We are told the actual work is 4 times the one above.
Thus;
Actual work = 4W = 4 × 911400 = 3,645,600 J
Now,
Molar mass of Glucose(C6H12O6) = 180 g/mol
We are given standard enthalpy of combustion = -1273.3 KJ/mol = -1273300
Moles of glucose = 3645600/1273300 = 2.863mol
Mass of glucose = 2.863 mol × 180 g/mol
Mass of glucose = 515.34 g
