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3 years ago

PLEASE HELP !

Chemistry

1 answer:

mario62 [17]3 years ago

4 0

Answer:

2L of water.

Explanation:

To know the volume of water to be added to the initial solution, first let us calculate the volume of the final solution. This is illustrated below:

Data obtained from the question:

Initial volume (V1) = 2L

Initial concentration (C1) = 6mol/L

Final concentration (C2) = 3mol/L

Final volume (V2) =?

Using the dilution formula, we can obtain the final volume of the stock as follow:

C1V1 =C2V2

6 x 2 = 3 x V2

Divide both side by 3

V2 = (6 x 2)/3

V2 = 4L.

The final volume of the solution is 4L.

To obtain the volume of water added, we shall determine the change in the volume of the solution. This is illustrated below:

Initial volume (V1) = 2L

Final volume (V2) = 4L

Change in volume = V2 – V1 = 4 – 2 = 2L.

Therefore, 2L of water must be added to the initial solution.

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The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3

9966 [12]

Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol

$k_2=2.3\times 10^8$

$T_2=280\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (280 + 273.15) K = 553.15 K

$T_2=553.15\ K$

$k_1=4.8\times 10^8$

$T_1=376\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (376 + 273.15) K = 649.15 K

$T_1=649.15\ K$

So,

$\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)$

$E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=22.87\ kJ/mol$

<u>The activation energy for this reaction = 23 kJ/mol.</u>

6 0

2 years ago

Which solution has the higher boiling point?

dedylja [7]

NaCl has higher boiling point than C₂H₆O₂ in H₂O.

<h3>What is molality?</h3>

The number of moles of solute in a solution equal to 1 kg or 1000 g of solvent is referred to as its molality.
The definition of molarity, on the other hand, is based on a certain volume of solution.
Mol/kg is a typical molality measurement unit in chemistry.

<h3>Calculation of boiling point:</h3>

When the non-volatile solute is dissolved in a solvent, the boiling point rises along with the molality (concentration) of the solute.

Given,

Mass of C₂H₆O₂ (solute) = 15.0 g

Mass of solvent (H₂O) = 0.50 kg

Molar mass of C₂H₆O₂ = 62 g/mol

Thus, the moles of solute = 15.0 g x 1.0 mol solute/62 g/mol

= 0.2419 mol

Therefore, molality(m) of C₂H₆O₂ (solute) = Amount of solute (mol)/ Mass of solvent

= 0.2419/0.50

= 0.4838 m

Similarly,

Given ,

Mass of NaCl (solute) = 15 g

Mass of solvent (H₂O) = 0.50 kg

Molar mass of NaCl (solute) = 58.44 g/mol

Thus, the moles of NaCl (solute) = 15.0 g x 1.0 mol solute/58.44

= 0.2566 mol

Therefore, molality(m) of NaCl (solute) = Amount of solute (mol)/ Mass of solvent

= 0.2566 mol/0.50 kg

= 0.51 m

Hence, the molality of NaCl (solute) is more than the molality of C₂H₆O₂(solute).

So, with an increase in the solute's concentration (molality), the boiling point rises.

Therefore, NaCl has higher boiling point than C₂H₆O₂ in H₂O.

Learn more about boiling point here:

brainly.com/question/24168079

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7 0

1 year ago

What fraction can be used in conversion of 32 m3 to mm3

zimovet [89]

The calculation is (Measurement in m³ x 1000³)

Therefore

Measurement in m³ x 1 000 000

Its not a fraction as there are 1000mm in a m, so to convert from m to mm you must multiply

5 0

2 years ago

Read 2 more answers

How many moles are in 2.8x10^24 atoms of silicon?

sergejj [24]

Answer:

4.6 mol Si

General Formulas and Concepts:

<u>Atomic Structure</u>

Reading a Periodic Table
Moles
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 2.8 × 10²⁴ atoms Si

[Solve] moles Si

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 2.8 \cdot 10^{24} \ atoms \ Si(\frac{1 \ mol \ Si}{6.022 \cdot 10^{23} \ atoms \ Si})$
[DA] Divide [Cancel like units]: $\displaystyle 4.64962 \ mol \ Si$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

4.64962 mol Si ≈ 4.6 mol Si

6 0

2 years ago

Convert 43 miles into feet

docker41 [41]

Answer:

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5 0

3 years ago