Answer is: the compound is B₂O₃.
ω(O) = 68.94% ÷ 100%.
ω(O) = 0.6894; percentage of oxygen in the compound.
ω(X) = 31.06% ÷ 100%.
ω(X) = 0.3106; percentage of unknown element in the compound.
If we take 69.7 grams of the compound:
M(compound) = 69.7 g/mol.
n(compound) = 69.7 g ÷ 69.7 g/mol.
n(compound) = 1 mol.
n(O) = (69.7 g · 0.6894) ÷ 16 g/mol.
n(O) = 3 mol.
M(compound) = n(O) · M(O) + n(X) · M(X).
n(X) = 1 mol ⇒ M(X) = 21.7 g/mol; there is no element with this molecular weight.
n(X) = 2 mol ⇒ M(X) = 10.85 g/mol; this element is boron (B).
Angles do not change they can only rotate in a circular motion if the angles in a 30°−60°−90° triangle after it is rotated clockwise at 45°.
<h3>What are the angles?</h3>
The angles are the distance between two lines that are attached at one point and they can vary in shapes like triangle and square or circle and rectangle.
The square has the handle of equality for all the sides rectangle has two opposite side angles equal and the circle It has only one angle triangle consisting of 3 angles which are fixed and cannot be changed.
Therefore, if the angles in a 30°−60°−90° triangle after it is rotated clockwise at 45°Angles do not change they can only rotate in a circular motion.
Learn more about angles, here:
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Answer:
(A). C6H5Br + Mg(in ether) -----------> C6H5MgBr.
(B). C6H5MgBr + O = C = O -----------> C6H5-COO^- Mg^+ Br.
(C). C6H5-COO^- Mg^+ Br + HCl --------> C6H5-COOH + Mg^+Br(OH).
PRODUCTS=> C6H5-COOH and Mg^+Br(OH).
Explanation:
A Grignard reagent is a reagent that/which is an organometallic compound that is R -Mg- X. The R = alkyl, vinyl or allyl and the X = halogens.
It must be noted that an important reaction of Grignard reagent is its reaction with compounds containing the Carbonyl that is -CO functional group and this kind of Reaction is known as a Grignard Reaction.
So, in this question we are told that;
=> "1-bromo-benzene andits subsequent reaction with solid carbon dioxide (CO2) followed by acidic workup (using HCl asthe acid). "
Thus;
(A). C6H5Br + Mg(in ether) -----------> C6H5MgBr.
(B). C6H5MgBr + O = C = O -----------> C6H5-COO^- Mg^+ Br.
(C). C6H5-COO^- Mg^+ Br + HCl --------> C6H5-COOH + Mg^+Br(OH).
For this, we use equations from the colligative properties of solutions specifically boiling point elevation and freezing point depression. The equations for these are expressed as:
ΔTb = kb m
where k is a boiling point elevation constant and m is the concentration in terms of molality
ΔTf = kf m
where k is a freezing point elevation <span>constant and m is the concentration in terms of molality
</span>
We use both expression to solve for the freezing point. For this case, concentration is the same. The equation will then be:
ΔTf = kf ( ΔTb / kb )
0-Tf = 1.86 (103.7 - 100 / 0.512 )
Tf = -13.4°C
