if all sides are unequal it is scalene and if one angle adds up to 90 while the rest are different youve proven it.
I believe the answer is letter C
Taking into account the reaction stoichiometry, 109.09 grams of Ag₂S₂O₃ are formed when 125 g AgBr reacts completely.
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
2 AgBr + Na₂S₂O₃ → Ag₂S₂O₃ + 2 NaBr
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- AgBr: 2 moles
- Na₂S₂O₃: 1 mole
- Ag₂S₂O₃: 1 mole
- NaBr: 2 moles
The molar mass of the compounds is:
- AgBr: 187.77 g/mole
- Na₂S₂O₃: 158 g/mole
- Ag₂S₂O₃: 327.74 g/mole
- NaBr: 102.9 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- AgBr: 2 moles ×187.77 g/mole= 375.54 grams
- Na₂S₂O₃: 1 mole ×158 g/mole= 158 grams
- Ag₂S₂O₃: 1 mole ×327.74 g/mole= 327.74 grams
- NaBr: 2 moles ×102.9 g/mole= 205.8 grams
<h3>Mass of Ag₂S₂O₃ formed</h3>
The following rule of three can be applied: if by reaction stoichiometry 375.54 grams of AgBr form 327.74 grams of Ag₂S₂O₃, 125 grams of AgBr form how much mass of Ag₂S₂O₃?

<u><em>mass of Ag₂S₂O₃= 109.09 grams</em></u>
Then, 109.09 grams of Ag₂S₂O₃ are formed when 125 g AgBr reacts completely.
Learn more about the reaction stoichiometry:
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Answer: a. 0.75m (NH4)3PO4 will have the lowest freezing point .
Explanation: Freezing point decreases as the concentration of the solute substance increases. Assuming the same solvent for all of them, for instance water.
∆T= i.Kf.b
∆T= freezing point depression
i= vant Hoff factor
Kf= molality
Assuming water to be the solvent for all Kf=1.86°C/M
VANT HOFF FACTORS :
For (NH4)3PO4
This has 3 ionic bonding and 1 covalent bonding making it 4 bond
Therefore i=4
For CaSO4
This has 1 metallic bond and 1 covalent bond making it 2 bond.
Therefore i=2
For LiCl
This has 1 metallic bond and 1 non metallic bond making it 2 bond.
Therefore i=2
For CH3OH
This has only 1 covalent bond.
Therefore i=1
MOLALITY:
(NH4)3PO4 = 0.75M
CaSO4= 1.0M
LiCl= 1.0M
CH3OH= 1.5M
FREEZING POINT DEPRESSION:
For (NH4)3PO4
∆T= 4×0.75×1.86=5.58°C
For CaSO4
∆T= 2×1.0×1.86=3.72°C
For LiCl
∆T = 2×1.0×1.86= 3.72°C
For CH3OH
∆T= 1×1.5×1.86=2.79°C
REMEMBER THE HIGHER THE FREEZING POINT DEPRESSION THE LOWER THE FREEZING POINT.
FREEZING POINT DEPRESSION IS THE CHANGE IN THE FREEZING POINT PROPORTIONAL TO THE AMOUNT OF SOLUTE ADDED THE THE SOLUTION.
THEREFORE THE ONE WITH THE LOWEST FREEZING POINT IS (NH4)3PO4