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2 years ago

How many grams of ethylene glycol (c2h6o2 must be added to 1.25 kg of water to produce a solution that freezes at -5.88 ∘c?

1 answer:

6 0

The freezing point depression is calculated through the equation,
ΔT = (kf) x m
where ΔT is the difference in temperature, kf is the freezing point depression constant (1.86°C/m), and m is the molality. Substituting the known values,
5.88 = (1.86)(m)
m is equal to 3.16m

Recall that molality is calculated through the equation,
molality = number of mols / kg of solvent
number of mols = (3.16)(1.25) = 3.95 moles
Then, we multiply the calculated amount in moles with the molar mass of ethylene glycol and the answer would be 244.9 g.

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A sample of gallium Bromide GaBr2,weighing 0.165 g was dissolved in water and treated with silver nitrate AgNO3, and resulting t

tresset_1 [31]

<u>Answer:</u> The percent gallium in gallium bromide is 30.30 %.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of gallium bromide = 0.165 g

Molar mass of titanium gallium bromide = 229.53 g/mol

Putting values in equation 1, we get:

$\text{Moles of gallium bromide}=\frac{0.165g}{229.53g/mol}=0.00072mol$

The chemical equation for the reaction of gallium bromide and silver nitrate follows:

$GaBr_2+2AgNO_3\rightarrow 2AgBr(s)+Ga(NO_3)_2$

By Stoichiometry of the reaction:

1 moles of gallium bromide produces 1 mole of gallium nitrate

So, 0.00072 moles of gallium bromide will produce = $\frac{1}{1}\times 0.00072=0.00072moles$ of gallium nitrate

Now, calculating the mass of gallium nitrate from equation 1, we get:

Molar mass of gallium nitrate = 193.73 g/mol

Moles of gallium nitrate = 0.00072 moles

Putting values in equation 1, we get:

$0.00072mol=\frac{\text{Mass of gallium nitrate}}{193.73g/mol}\\\\\text{Mass of gallium nitrate}=0.139g$

Calculating the mass of gallium in the reaction, we use unitary method:

In 1 mole of gallium nitrate, 1 mole of gallium atom is present.

In 193.73 grams of gallium nitrate, 69.72 g of gallium atom is present.

So, in 0.139 grams of gallium nitrate, the mass of gallium present will be = $\frac{69.72}{193.73}\times 0.139=g$

To calculate the percentage composition of gallium in gallium bromide, we use the equation:

$\%\text{ composition of gallium}=\frac{\text{Mass of gallium}}{\text{Mass of gallium bromide}}\times 100$

Mass of gallium bromide = 0.165 g

Mass of gallium = 0.050 g

Putting values in above equation, we get:

$\%\text{ composition of gallium}=\frac{0.050g}{0.165g}\times 100=30.30\%$

Hence, the percent gallium in gallium bromide is 30.30 %.

3 0

3 years ago

A gas sample occupies 350.0 mL at 546 mm Hg. What volume does the gas occupy at 652 mm Hg?

leva [86]

Answer:

293.1 mL.

Explanation:

Boyle's law states that: at a constant temperature the pressure of a given mass of an ideal gas is inversely proportional to its volume.

It can be expressed as: <em>P₁V₁ = P₂V₂,</em>

P₁ = 546.0 mm Hg, V₁ = 350.0 mL.

P₂ = 652.0 mm Hg, V₂ = ??? mL.

8 0

2 years ago

Is P2I4 or Diphosphorus tetraiodide covalent or ionic?

elena55 [62]

Diphosphorus tetraiodide is a covalent compound.

It has low melting point as compared to ionic compounds

It is a rare compound where the oxidation state of Phosphorous is +2.

It is also termed as subhalide of phosphorous.

5 0

2 years ago

Read 2 more answers

Write a balanced half-reaction for the product that forms at each electrode in the aqueous electrolysis of the following salts:(

likoan [24]

Balanced Half reactions are:

At anode 2 $Cl^{-}$ ==> Cl₂+ $2e^{-}$ + H₂O ==> $ClO^{-}$ + 2 $H^{+}$ + $Cl^{-}$

At Cathode: 2 $H^{+}$ + $2e^{-}$ ==> H₂

Since the question states that you are using an aqueous solution of MnCl₂, so ions will have present are, H₂O, $H^{+}$ , $Mn^{2+}$ and $Cl^{-}$

Now at Anode reaction will occur as given:

2 $Cl^{-}$ ==> Cl₂+ $2e^{-}$ + H₂O ==> $ClO^{-}$ + 2 $H^{+}$ + $Cl^{-}$ (will occur)

At Cathode:

2 $H^{+}$ + $2e^{-}$ ==> H₂ (will occur)

At Cathode:

$Mn^{2+}$ + $2e^{-}$ ==> Mn (This reaction will not occur)

The deposition of solid Mn will not occur because in aqueous solution, $H^{+}$ will be reduced before $Mn^{2+}$ .

The reduction potentials for $H^{+}$ is zero whereas reduction potential for $Mn^{2+}$ is - 1.18V.

The reduction potential of a species is its tendency to gain electrons and get reduced. It is measured in millivolts or volts. Larger positive values of reduction potential are indicative of a greater tendency to get reduced.

To learn more about the half reaction please click on the link brainly.com/question/13186640

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8 0

1 year ago

es the molecule defective leading to sickle cell anemia. Predict whether the following hypothetical change would or would not ha

Pachacha [2.7K]

<h2><u>Full Question:</u></h2>

In hemoglobin, a single amino acid change at position 6 from Glu to Val has major consequences on hemoglobin structure that makes the molecule defective leading to sickle cell anemia. Predict whether the following hypothetical change would or would not have a major effect at position 6. Briefly explain (1-2 sentences). Glu to Leu Hint: Look at the structures of the R groups and consider their chemical properties

<h2><u>Answer:</u></h2>

The structure of the haemoglobin, hence the RBC won't be same as normal.

<h3><u>Explanation:</u></h3>

Both the leucine and glutamic acid are alpha amino acids which have an alpha carboxylic acid group and an alpha amino group. The variable in case of glutamic acid is propyl acid while the variable in case of leucine is isobutyl.

The glutamic acid is the normal amino acid of the 6th position of Beta chain of hemoglobin. Its an acid group, so can form bonds with another base inside the haemoglobin, or can form other hydrogen bonds. But the isobutyl group is an alkyl group. So it doesn't have that much effect in the recovering the structure, and sickle cell anemia prevails.

6 0

3 years ago