Can someone Help please !!!!!

Chemistry

1 answer:

Georgia [21]3 years ago

7 0

The correct answer is C.

On the left:
2 carbon+ 2 oxygen+ 2 oxygen

On the right:
2 carbon+ 4 oxygen (remember to multiply the subscript by the coefficient)

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g Sucrose (C12H22O11), a nonionic solute, dissolves in water (normal freezing/melting point 0.0°C) to form a solution. If some u

DIA [1.3K]

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

4 0

4 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.