This problem has been solved! See the answer Determine Um (mode ), average U, and Urms for a group of ten automobiles clocked by
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Answer : The correct option is, Only Student B
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is,
As we know that nitrogen has '5' valence electrons and hydrogen has '1' valence electron.
Therefore, the total number of valence electrons in = 5 + 3(1) = 8
According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.
The Lewis dot structure of student A is wrong because there is a coordinate bond present between the nitrogen and hydrogen is not covalent.
Thus, the correct Lewis-dot structure of is shown by the student B.
<u>Answer:</u> The volume of CO formed is 254.43 L.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
Given mass of = 444 g
Molar mass of = 170.73 g/mol
Putting values in above equation, we get:
For the given chemical reaction:
By stoichiometry of the reaction:
1 mole of nickel tetracarbonyl produces 4 moles of carbon monoxide.
So, 2.60 moles of nickel tetracarbonyl will produce = of carbon monoxide.
Now, to calculate the volume of the gas, we use ideal gas equation, which is:
PV = nRT
where,
P = Pressure of the gas = 752 torr
V = Volume of the gas = ? L
n = Number of moles of gas = 10.4 mol
R = Gas constant =
T = Temperature of the gas =
Putting values in above equation, we get:
Hence, the volume of CO formed is 254.43 L.
Answer:
Explanation:
Hello!
In this case, since the chemical reaction between copper and nitric acid is:
By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:
However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:
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