Answer:
C
Explanation:
electronic waves transmits energy but mechanical waves require a medium in order to transport their energy from bgg one place to another.
Answer:
1255.4L
Explanation:
Given parameters:
P₁ = 928kpa
T₁ = 129°C
V₁ = 569L
P₂ = 319kpa
T₂ = 32°C
Unknown:
V₂ = ?
Solution:
The combined gas law application to this problem can help us solve it. It is mathematically expressed as;

P, V and T are pressure, volume and temperature
where 1 and 2 are initial and final states.
Now,
take the units to the appropriate ones;
kpa to atm, °C to K
P₂ = 319kpa in atm gives 3.15atm
P₁ = 928kpa gives 9.16atm
T₂ = 32°C gives 273 + 32 = 305K
T₁ = 129°C gives 129 + 273 = 402K
Input the values in the equation and solve for V₂;

V₂ = 1255.4L
Answer:
Aluminium.
Explanation:
The above electronic configuration can be written in a simplified form as shown below:
1s² 2s²2p⁶ 3s²3p¹
Next, we shall determine the number of electrons in the atom of the element as follow:
Number electron = 2 + 2 + 6 + 2 + 1
Number of electron = 13
Next, we shall determine the number of protons.
Since the element is in its neutral state,
The number of electrons and protons are equal i.e
Proton = Electron
Number of electron = 13
Proton = Electron = 13
Proton = 13
Next, we shall determine the atomic number of the element.
The atomic number of an element is simply the number of protons in the atom of the element i.e
Atomic number = proton number
Proton = 13
Atomic number = 13
Comparing the atomic number of the element with those in the periodic table, the element with the above electronic configuration is aluminium since no two elements have the same atomic number.
Explanation:
Mg(s) + Cr(C2H3O2)3 (aq)
Overall, balanced molecular equation
Mg(s) + Cr(C2H3O2)3(aq) --> Mg(C2H3O2)3(aq) + Cr(s)
To identify if an element has been reduced or oxidized, the oxidation number is observed in both the reactant and product phase.
An increase in oxidation number denotes that the element has been oxidized.
A decrease in oxidation number denotes that the element has been reduced.
Oxidation number of Mg:
Reactant - 0
Product - +3
Oxidation number of Cr:
Reactant - +3
Product - 0
Note: C2H3O2 is actually acetate ion; CH3COO- The oxidatioon number of C, H and O do not change.
Oxidized : Mg
Reduced : Cr
Answer:
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Explanation:
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