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1 year ago

How many g i2 should be added to 750g ccl4 to prepare a 0.200m solution?

2 answers:

6 0

There are a number of ways to express concentration of a solution. This includes molality. Molality is expressed as the number of moles of solute per mass of the solvent. We calculate as follows:

0.200 mol I2 / kg CCl4 ( .750 kg CCl4 ) ( 253.809 g I2 / mol I2) = 38.07 g I2 needed

Hope this helps.

tamaranim1 [39]1 year ago

3 0

Answer:

Mass of I2 required = 38.1 g

Explanation:

Given:

Mass of CCl4 = 750 g = 0.750 kg

Molality of the solution = 0.200 m

To determine:

Mass of I2

Formula:

Molality of a solution is expressed as the ratio of the amount of solute per kg of solvent

$Molality = \frac{moles\ of\ solute}{kg\ solvent} \\$

Here the solute is I2 and solvent is CCl4

$Moles\ of\ I2 = molality*mass\ solvent = 0.200*0.750= 0.150 moles\\\\Molar\ mass\ of\ I2 = 253.8 g/mol\\\\Mass\ of\ I2 = moles*molar\ mass = 0.150*253.8 = 38.1 g$

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A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 6

JulijaS [17]

Specific heat is the amount of heat absorb or released by a substance to change the temperature to one degree Celsius. To determine the specific heat, we use the expression for the heat absorbed by the system. Heat gained or absorbed in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:
Heat = mC(T2-T1)
By substituting the given values, we can calculate for C which is the specific heat of the material.
2510 J = .158 kg ( 1000 g / 1 kg) (C) ( 61.0 - 32.0 °C)C = 0.5478 J / g °C

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1 year ago

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Only one isotope of this element occurs in nature. one atom of this isotope has a mass of 9.123 ✕ 10-23 g. identify the element

o-na [289]

The atom has only one isotope which means 100 % of same atom is present in nature. The atomic mass of an element is the number of times an atom of that element is heavier than an atom of carbon taken as 12. Mass of one atom of that isotope is 9.123 ✕ 10⁻²³ g, so mass of one mole of atom that is Avogadro's number of atom is 6.023 X 10²³ X 9.123 X 10⁻²³ g=54.94 g = 55 g (approximate).

So, the atom having atomic mass 55 will be Cesium (Cs). Only one isotope of Cesium is stable in nature.

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1 year ago

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175 mL of Cl2 gas is held in a flexible vessel at

Gnoma [55]

Answer:

V₂ = 15.6 L

Explanation:

Given data:

Initial volume = 175 mL (0.175 L)

Initial pressure = 1 atm

Initial temperature = 273 K

Final temperature = -5°C (-5+273 = 268 K)

Final volume = ?

Final pressure = 1.16 kpa (1.16/101=0.011 atm)

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ = 1 atm × 0.175 L × 268 K / 273 K × 0.011 atm

V₂ = 46.9 L / 3.003

V₂ = 15.6 L

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1 year ago

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I NEED HELP ASAP, WILL MARK BRAINLEST!

Andre45 [30]

Answer:

1. 90%

2. 217.4 g O₂

3. 95.0%

4. Trial 2 ratios

Explanation:

Original: SiCl₄ + O₂ → SiO₂ + Cl₂

Balanced: SiCl₄ + O₂ → SiO₂ + 2Cl₂

Trial SiCl₄ O₂ SiO₂

1 120 g 240 g 38.2 g

2 75 g 50 g 25.2 g

Percentage yield for trial 1

We need to get actual yield (38.2 g) and theoretical yield, in grams.

Mass to moles:

molar mass SiCl₄: 28.09 + 4(35.45) = 169.9 g/mol

120 g SiCl₄ x 1 mol/169.9 g = .706 mol SiCl₄

Moles to moles:

For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

.706 mol SiCl₄ = .706 mol SiO₂

Moles to mass:

molar mass SiO₂: 28.09 + 2(16.00) = 60.09 g/mol

.706 mol SiO₂ x 60.09g/mol = 42.44 g SiO₂

Theoretical yield:

actual/theoretical x 100

38.2 / 42.44 = .900 = 90.0% yield

Leftover reactant for trial 1

We know oxygen is the excess reactant.

Mass to moles:

molar mass O₂ = 32.00 g/mol

240 g O₂ x 1 mol/32.00 g = 7.5 mol O₂

We used .706 mol SiO₂, so we also used .706 mol O₂.

7.5 - .706 = 6.8 moles left over

Moles to mass:

6.8 mol O₂ x 32.00g/mol = 217.4 g O₂

Percentage yield for trial 2

Mass to moles:

molar mass SiCl₄: 169.9 g/mol

75 g SiCl₄ x 1 mol/169.9 g = .441 mol SiCl₄

Moles to moles:

For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

.441 mol SiCl₄ = .441 mol SiO₂

Moles to mass:

molar mass SiO₂: 60.09 g/mol

.441 mol SiO₂ x 60.09g/mol = 26.5 g SiO₂

Theoretical yield:

actual/theoretical x 100

25.2 / 26.5 = .950 = 95.0% yield

Because the percentage yield of trial 2 is higher than that of trial 1, we know that the ratio of reactants in trial 2 is more efficient! We got a result closer to our theoretical yield.

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2 years ago

How many acidic protons are there in 0.6137 g of KHP?

andreyandreev [35.5K]

0.6137 g of KHP contains 1.086 × 10^21 acidic protons.

Number of moles of KHP = mass of KHP/molar mass of KHP

Molar mass of KHP = 204.22 g/mol

Mass of KHP = 0.6137 g

Number of moles of KHP = 0.6137 g/204.22 g/mol = 0.003 moles of KHP

Now, 1 each molecule of KHP contains 1 acidic proton.

For 0.003 moles of KHP there are; 0.003 × 1 × NA

Where NA is Avogadro's number.

So; 0.003 moles of KHP contains 0.003 × 1 × 6.02 × 10^23

= 1.086 × 10^21 acidic protons.

Learn more: brainly.com/question/16672114

7 0

1 year ago