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3 years ago

What is the freezing point of an aqueous solution made by dissolving 0.118 mole of MgCl2 in 100.0 g of water? Kf = 1.86 oC/m

Chemistry

1 answer:

Sauron [17]3 years ago

8 0

Answer:

Freezing point of solution is -4.39°C

Explanation:

Freezing point depression's formula:

ΔT = Kf . m . i

ΔT = Freezing point of pure solvent - Freezing point of solution

0°C - X = 1.86°C/m . m . 2

i = Van't Hoff factor, ions dissolved in solution

MgCl₂ → Mg²⁺ + 2Cl⁻ i = 3 → 1 mol of Mg and 2 mol of chloride

Let's think the molality (mol/kg)

First of all, let's convert the mass of solvent from g to kg

100 g / 1000 = 0.1 kg

Molality (mol/kg) = 0.118 mol / 0.1 kg = 1.18 m

0°C - X = 1.86°C/m . 1.18 m . 2

X = - 4.39°C

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Which equation represents a transuranium

Maksim231197 [3]

Answer:

1) Since you have not provided the equations to select the right one, I am going to explain you the relevant facts that are used to solve this question.

2) The transuranium elements are the chemiical elements with atomic number greater than that of the uranium.

The atomic number of uranium is 92. So, the transuranium elements are the elements with atomic number 93 or greater.

This are some of the transuranium elements:

Neptunio - 93

Plutonium - 94

Americium - 95

Curium - 96

Berkelium - 97

Californium - 98

Einstenium - 99

And so all the known elements (the last one is the 118).

3) In a nuclear reaction the total mass number ( shown as superscript to the left of the symbol) and total atomic number (shown as subscript to the left of the symbol) are conserved.

4) Beta decay is the release of a beta particle, which is an electron (considered massles and with charge - 1). So, the beta decay is represented with the symbol:

β, which means 0 mass and charge - 1.

-1

5) This is, then, an example of a β decay equation for one transuranium element:

239 239 0

Np → Pu + β

93 94 -1

As you see 239 = 239 + 0 and 93 = 94 - 1, showing that the total mass number ( shown as superscript to the left of the symbol) and the total atomic number (shown as subscript to the left of the symbol) are conserved.

Explanation:

3 0

3 years ago

Assuming that an acetic acid solution is 12% by mass and that the density of the solution is 1.00 g/mL, what volume of 1 M NaOH

Doss [256]

Explanation:

Let us assume that total mass of the solution is 100 g. And, as it is given that acetic acid solution is 12% by mass which means that mass of acetic acid is 12 g and 88 g is the water.

Now, calculate the number of moles of acetic acid as its molar mass is 60 g/mol.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{12 g}{60 g/mol}$

= 0.2 mol

Molarity of acetic acid is calculated as follows.

Density = $\frac{mass}{volume}$

1 g/ml = $\frac{100 g}{volume}$

volume = 100 ml

Hence, molarity = $\frac{\text{no. of moles}}{volume}$

= $\frac{0.2 mol}{0.1 L}$

= 2 mol/l

As reaction equation for the given reaction is as follows.

$NaOH + CH_{3}COOH \rightarrow CH_{3}COONa + H_{2}O$

So, moles of NaOH = moles of acetic acid

Let us suppose that moles of NaOH are "x".

$x \times 1 M = 10 mL \times 2 M$ (as 1 L = 1000 ml)

x = 20 L

Thus, we can conclude that volume of NaOH required is 20 ml.

6 0

3 years ago

If the initial temperature of a movable cylinder was 50 degrees Celsius

slega [8]

Answer:

8.45 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 50 °C

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

Initial temperature (T₁) = 323 K

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 273 k

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

2 × 5 / 323 = 1 × V₂ / 273

10 / 323 = V₂ / 273

Cross multiply

323 × V₂ = 10 × 273

323 × V₂ = 2730

Divide both side by 323

V₂ = 2730 / 323

V₂ = 8.45 L

Thus, the new volume is 8.45 L

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3 years ago

How many moles of Cu2S can be formed from 11 moles of HCl?

natima [27]

Answer:

Explanation:

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2 years ago

If 3.289 x 10^23 atoms of potassium react with excess water, how many grams of hydrogen gas would be produced?

GarryVolchara [31]

Answer:

The amount in grams of hydrogen gas produced is 0.551 grams

Explanation:

The parameters given are;

Number of atoms of potassium, aₙ = 3.289 × 10²³ atoms

Chemical equation for the reaction is given as follows;

2K + 2H₂O $\rightarrow$ KOH + H₂

Avogadro's number, $N_A$ , regarding the number of molecules or atom per mole is given s follows;

$N_A$ = 6.02 × 10²³ atoms/mole

Therefore;

The number of moles of potassium present = 3.289 × 10²³/(6.02 × 10²³) = 0.546 moles

2 moles of potassium produces one mole of hydrogen gas, therefore;

1 moles of potassium produces 1/2 mole of hydrogen gas, and 0.546 moles of potassium will produce 0.546/2 moles of hydrogen which is 0.273 moles of hydrogen gas

The molar mass of hydrogen gas = 2.016 grams

Therefore, 0.273 moles will have a mass of 0.273×2.016 = 0.551 grams.

The amount in grams of hydrogen gas produced = 0.551 grams.

8 0

3 years ago