The climate would become colder...
We find the weight of the empirical formula:
12.0107 + 2 x 1.00794 + 15.9994
= 30.03
Now, we divide the molecular weight by the weight of the empirical formula to find the number of times the empirical formula repeats:
90.09 / 30.03
= 3
The formula is 3(CH₂O)
C₃H₆O₃
Answer:
15.4 g of sucrose
Explanation:
Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m
0.56°C / 1.86 m/°C = m → 0.301 mol/kg
m → molality (moles of solute in 1kg of solvent)
Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg
0.301 mol/kg . 0.150kg = 0.045 moles.
We determine the mass of sucrose, by the molar mass:
0.045 mol . 342 g/1mol = 15.4 g
An element or compound which occurs naturally in the earth is a mineral
The bond angles a and b are 120° respectively. The bond angle c is 111.4° .while the bond angle d is 120°. The bond angles e and f are 120° respectively.
In the carbonate ion, all the bond angles and bond lengths are equal hence three equivalent resonance structures can be drawn for the ion. All the bond angles, ( a and b) in carbonate ion all have bond angle of 120°.
The bond angle marked c in OCCl2 has a bond angle 111.4°, the bond angle marked d in the compound has the bond angle, 120°.
There are three bond angles present in the nitrate (NO3-) ion. Three resonance structures contribute to this bond. Based on these structures, the bond angles e and f in the molecule is 120°.
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