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4 months ago

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(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak

e place and before equilibrium is established), was 5.56. On the following graph, plot the points representing the initial concentrations of all three gases. Label each point with the formula of the gas.

1 answer:

KiRa [2.9K]4 months ago

7 0

Response:

Here's my calculation

Clarification:

Assume the starting concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We need to determine the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's compile all the information in one location.

H₂ + I₂ ⇌ 2HI

I/mol·L⁻¹: 0.30 0.15 x

2. Calculate the concentration of HI

$Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}$

3. Plot the initial values

The graph below visualizes the initial concentrations as plotted on the vertical axis.

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Find the molarity of 750 ml solution containing 346 g of potassium nitrate

lorasvet [2795]

For KNO₃, the mass is 346g. The molar mass can be computed as (39.098) + (14) + (15.99*3), which results in 101.068 gmol⁻¹. The volume of the solution is given as 750ml, equivalent to 0.75dm³. The formula for molarity is (mass of solute/molar mass of solute)*(1/volume of solution in dm³). Accordingly, molarity = (346/101.068)*(1/0.75), yielding 4.56 mol dm⁻³.

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2 months ago

Ibuprofen, a headache remedy, contains 75.69% C, 8.80% H, and 15.51% O by mass and has a molar mass of 206 g/mol. Express your a

castortr0y [3046]

Answer:

A) The molecular formula for ibuprofen is $C_{13}H_{18}O_2$

B) The molecular formula for Cadaverine is $C_{5}H_{14}N_2$

C) The molecular formula for Epinephrine is $C_9H_{13}O_3N_1$

Explanation:

Element percentage in a compound:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of element}}\times 100$

A) The composition of ibuprofen, used for headaches, consists of 75.69% carbon, 8.80% hydrogen, and 15.51% oxygen by weight.

Ibuprofen has a molar mass of 206 g/mol.

The proposed molecular formula for ibuprofen is = $C_xH_yO_z$

Count of carbon atoms in one ibuprofen molecule;

$75.69\%=\frac{x\times 12 g/mol}{206 g/mol}\times 100$

$x=\frac{75.69\times 206 g/mol}{100\times 12 g/mol}=12.99\approx 13$

Count of hydrogen atoms in one ibuprofen molecule;

$8.80\%=\frac{y\times 1 g/mol}{206 g/mol}\times 100$

$y=\frac{8.80\times 206 g/mol}{100\times 1 g/mol}=18.12\approx 18$

Count of oxygen atoms in one ibuprofen molecule;

$15.51\%=\frac{z\times 16 g/mol}{206 g/mol}\times 100$

$z=\frac{15.51\times 206 g/mol}{100\times 16 g/mol}=1.99\approx 2$

Molecular formula for ibuprofen:

= $C_xH_yO_z= C_{13}H_{18}O_2$

B) Cadaverine consists of 58.55% carbon, 13.81% hydrogen, and 27.40% nitrogen by weight

Cadaverine has a molar mass of 102.2 g/mol.

The proposed molecular formula for Cadaverine is = $C_xH_yN_z$

Count of carbon atoms in one Cadaverine molecule;

$58.55\%=\frac{x\times 12 g/mol}{102.2 g/mol}\times 100$

$x=\frac{58.55\times 102.2 g/mol}{100\times 12 g/mol}=4.98\approx 5$

Count of hydrogen atoms in one Cadaverine molecule;

$13.81\%=\frac{y\times 1 g/mol}{102.2 g/mol}\times 100$

$y=\frac{13.81\times 102.2 g/mol}{100\times 1 g/mol}=14.11\approx 14$

Count of nitrogen atoms in one Cadaverine molecule;

$27.40\%=\frac{z\times 14 g/mol}{102.2 g/mol}\times 100$

$z=\frac{27.40\times 102.2 g/mol}{100\times 14 g/mol}=2.00\approx 2$

Molecular formula for Cadaverine:

= $C_xH_yN_z= C_{5}H_{14}N_2$

C) Epinephrine includes 59.0% carbon, 7.1% hydrogen, 26.2% oxygen, and 7.7% nitrogen by weight

Epinephrine has a molar mass of 180 g/mol.

The proposed molecular formula for Epinephrine is = $C_xH_yO_zN_w$

Count of carbon atoms in one Epinephrine molecule;

$59.0\%=\frac{x\times 12 g/mol}{180 g/mol}\times 100$

$x=\frac{59.0\times 180 g/mol}{100\times 12 g/mol}=8.85\approx 9$

Count of hydrogen atoms in one Epinephrine molecule;

$7.1\%=\frac{y\times 1 g/mol}{180 g/mol}\times 100$

$y=\frac{7.1\times 180 g/mol}{100\times 1 g/mol}=12.78\approx 13$

Count of oxygen atoms in one Epinephrine molecule;

$26.2\%=\frac{z\times 16 g/mol}{180 g/mol}\times 100$

$z=\frac{26.2\times 180 g/mol}{100\times 16 g/mol}=2.94\approx 3$

Count of nitrogen atoms in one Epinephrine molecule;

$7.7\%=\frac{w\times 14 g/mol}{180 g/mol}\times 100$

$w=\frac{7.7\times 180 g/mol}{100\times 14 g/mol}=0.99\approx 1$

Molecular formula for Epinephrine:

= $C_xH_yO_zN_w= C_9H_{13}O_3N_1$

7 0

4 months ago