The answer is b because that’s it
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
Single-celled organisms<span> which use asexual reproduction can </span>do<span> so very rapidly simply by </span>dividing<span> into two equal halves. This is called binary fission. In yeasts the </span>cell<span> does not </span>divide<span> equally in two halves; instead, there is a large mother </span>cell<span> and a smaller daughter </span>cell<span>. This is called budding.</span>
Acid is 1. a substance that increases the hydrogen ion concentration of a solution
<span>2. removes hydroxide ions because of the tendency for H+ to combine with OH-</span>