0.3268 moles of PC15 can be produced from 58.0 g of Cl₂ (and excess
P4)
<h3>How to calculate moles?</h3>
The balanced chemical equation is

The mass of clorine is m(
) = 58.0 g
The amount of clorine is n(
) = m(
)/M(
) = 58/70.906 = 0.817 mol
The stoichiometric reaction,shows that
10 moles of
yield 4 moles of
;
0.817 of
yield x moles of 
n(
) = 4*0.817/10 = 0.3268 mol
To know more about stoichiometric reaction, refer:
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Percent error can be calculated by the difference of the theoretical value and the measured value divided by the theoretical value multiplied by 100 percent.
% error = 27.26 - 27.2 / 27.26 x100
% error = 0.22%
A value close to zero would mean that the measured value is more or less near the actual value.
I'm going to say about 1-2 hours until it will completely dry. It depends on the material of the clothing.
Answer: The correct answer is "B" two bonding domains(or bonding pairs) or two non bonding domains(or lone pairs)
Explanation:
Molecular geometry/structure is a three dimensional shape of a molecule. It is basically an arrangement of atoms in a molecule.It is determined by the central atom, its surrounding atoms and electron pairs.According to VSEPR theory, there are 5 basic shapes of a molecule: linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral.
A)Four bonding domains and zero non bonding domains: shape is tetrahedral and bond angle is 109.5°
B)Two bonding domains and two non bonding domains(lone pairs): shape is bent and bond angle is 104.5°
C)Three bonding domains and one non bonding domain: shape is trigonal pyramidal and bond angle is 107°
D)Two bonding domain and zero non bonding domain: shape is linear and bond angle is 107°
E)Two bonding domain and one non bonding domain: bent shape and bond angle is 120°
F)Three bonding domains and zero nonbonding domain: shape is trigonal planar and bond angle is 120°
Hence Two bonding domains and two non bonding domains have the smallest bond angle.
Here we have to get the product between the reaction of butane-1-amine with methyl iodide (CH₃I).
The reaction between 1 mole of butan-1-amine and 1 mole of methyl iodide produces Methyl-butamine which is a secondary amine.
However, In presence of 2 moles of methyl iodide the reaction proceed to N, N-di-methylbutamine. The reaction is shown in the figure.
This is one of the effective reaction method to generate secondary and tertiary amine from primary amine.
The primary amine reacts with alkyl iodide to form secondary to tertiary amine. The final product depends upon the quantity of the alkyl iodide present in the reaction.