НСО, 11. ___ Nacio, > --

Determine the freezing point of an aqueous solution containing 10.50 g of magnesium bromide in 200.0 g of water.

Rudiy27

For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:

ΔTf = (Kf)(m)(i)
where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3

Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol

m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg

For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf

Tf = -1.59 celsius

5 0

3 years ago

How many ounces are in 0.442 L? Use dimensional analysis to

vovangra [49]

Answer:

14.945798 in us fluid ounce

Explanation:

3 0

3 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.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

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

8 0

3 years ago

Copper metal has a specific heat of 0.385 J/goC. Calculate the amount of heat required to raise the temperature of 22.8 g of Cu

Liula [17]

Answer is equal to 7.51 kJ

Q=mc(delta)T

4 0

2 years ago

Anything in red is the question

laila [671]

Answer:

Question 19: the three are molecular compounds.

Question 20: CuSO₄.5H₂O

Explanation:

Question 19.

C₂H₄

HF

H₂O₂

All of them are the combination of two kinds of different atoms in fixed proportions.

C₂H₄: two carbon atoms per four hydrogen atoms

HF: one hydrogen atom per one fluorine atom

H₂O₂: two hydrogen atoms per two oxygent atoms

Thus, they all meet the definition of compund: a pure substance formed by two or more different elements with a definite composition.

Molecular compounds are formed by covalent bonds and ionic compounds are formed by ionic bonds.

Two non-metal elements, like H-F, C - C, C - H, H-O, H - H, and O - O will share electrons forming covalent bonds to complete their valence shell. Thus, the three compounds are molecular and not ionic.

Question 20. Formula of copper(II) sulfate hydrate with 36.0% water.

Copper(II) sulfate is CuSO₄. Its molar mass is 159.609g/mol

Water is H₂O. Its molar mass is 18.015g/mol

Calling x the number of water molecules in the hydrate, the percentage of water is:

$\dfrac{18.015x}{18.015x+159.609}=36\%\\ \\ \\ \dfrac{18.015x}{18.015x+159.609}=0.36$

From which we can solve for x:

$18.015x=6.4854x+57.45924\\ \\ 11.5296x=57.45924\\ \\ x\approx4.98\approx5$

Thus, there are 5 molecules of water per each unit of CuSO₄, and the formula is:

CuSO₄.5H₂O

4 0

3 years ago

НСО, 11. ___ Nacio, > -- _NaCl +