<span>Xenon has 8 valence electrons, each hydrogen atom has 1, therefore you have 8+4=12 valence electrons. 8 electrons are used to make 4 bonds between the Xe atom and the 4 hydrogen atoms. That leaves 12-8=4 left over electrons, which will occur in two lone pairs of electrons on the center xenon atom. There are 4 bonding pairs and two lone pairs of electrons and they will be arranged in an octahedral arrangement around the center atom, with the lone pairs across the molecule (180 degrees from each other). The resulting molecular structure will be square planar.</span><span>
8+4=12 valence electrons. 8 electrons are used to make 4 bonds between the Xe atom and the 4 hydrogen atoms. That leaves 12-8=4 left over electrons, which will occur in two lone pairs of electrons on the center xenon atom. There are 4 bonding pairs and two lone pairs of electrons and they will be arranged in an octahedral arrangement around the center atom, with the lone pairs across the molecule . The resulting structure is square planar.</span>
Answer:
Yes
Explanation:
Because they support consumers. This is usually a green plant, because plants can make their own food by photosynthesis .
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Answer:
1.552 moles
Explanation:
First, we'll begin by writing a balanced equation for the reaction showing how C8H18 is burn in air to produce CO2.
This is illustrated below:
2C8H18 + 25O2 -> 16CO2 + 18H2O
Next, let us calculate the number of mole of C8H18 present in 22.1g of C8H18. This is illustrated below:
Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol
Mass of C8H18 = 22.1g
Mole of C8H18 =..?
Number of mole = Mass /Molar Mass
Mole of C8H18 = 22.1/144
Mole of C8H18 = 0.194 mole
From the balanced equation above,
2 moles of C8H18 produced 16 moles of CO2.
Therefore, 0.194 mole of C8H18 will produce = (0.194x16)/2 = 1.552 moles of CO2.
Therefore, 1.552 moles of CO2 are emitted into the atmosphere when 22.1 g C8H18 is burned
We can write the balanced equation for the galvanic cell by using the oxidation and reduction half-reactions. In the anode, Ni(s) is oxidized and produces aqueous Ni2+:
Ni(s) → Ni2+
while in the cathode, Cu2+ is reduced and deposits copper:
Cu2+ → Cu(s)
We now balance the charge of each reaction by adding electrons to the side of the equation with the greater charge:
Ni(s) → Ni2+ + 2 e-
Cu2+ + 2 e- → Cu(s)
Finally, we add the half-reactions to obtain the overall balanced equation for the galvanic cell:
Cu2+(aq) + Ni(s) → Ni2+(aq) + Cu(s)