2.) Determine the concentration of citric acid if a 20.00 mL sample is titrated with 29.06 mL of

Solution A is 0.44 M and reacts with 0.11 M of solution B. Assume that the value of x is 0, the value of y is 1, and r is 1.07 ×

xz_007 [3.2K]

Answer:

K, the rate constant = 9.73 × 10^(-1)/s

Explanation:

r = K × [A]^x × [B]^y

r = Rate = 1.07 × 10^(-1)/s

K = Rate constant

A and B = Concentration in mol/dm^-3

A = 0.44M

B = 0.11M

x = Order of reaction with respect to A = 0

y = Order of reaction with respect to B = 1

Solving, we get

r/([A]^x × [B]^y) = K

K = 1.07 × 10^(-1)/s/(0.44^0 × 0.11^1)= 0.9727

K = 0.9727

7 0

3 years ago

Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C

ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)

The enthalpy of reaction (1) is given by:

$\Delta H = \Delta H_{r} - \Delta H_{p}$ (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond

7O₂: 7 moles of 1 O=O bond

4CO₂: 4 moles of 2 C=O bonds

6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

$\Delta H = \Delta H_{r} - \Delta H_{p}$

$\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})$

$\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})$

$\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol$

$\Delta H = -2860 kJ/mol$

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

I hope it helps you!

7 0

3 years ago

Four springs are stretched to the same distance from the equilibrium position. The spring constants are listed in the table. A 2

svp [43]

Answer:

The correct option is;

X, W, Y, Z

Explanation:

The parameters given are;

Spring (S), Spring Constant (N/m)

W, 24

X, 35

Y, 22

Z, 15

The equation for elastic potential energy, $E_e$ , is $E_e = 0.5 \times k \times x^2$

The above equation can also be written as $E_e =\dfrac{1}{2} \times k \times x^2$

Where:

k = The spring constant in (N/m)

x = The spring extension

Therefore, since the elastic potential energy, $E_e$ , of the spring is directly proportional to the spring constant, k, we have the springs with higher spring constant will have higher elastic potential energy, $E_e$ , therefore the correct order is as follows;

X > W > Y > Z

9 0

3 years ago

Which has a greater volume, 10 grams of water or 10 grams of acetone?

WINSTONCH [101]

Density is defined as the ratio of mass to the volume.

Density = $\frac{Mass}{Volume}$ (1)

Mass of water = 10 grams

Mass of acetone = 10 grams

Density of water = 1 $\frac{g}{cm^{3}}$

Density of acetone = 0.7857 $\frac{g}{cm^{3}}$

Put the value of density of water and its mass in equation (1)

1 $\frac{g}{cm^{3}}$ = $\frac{10 g}{volume}$

Volume of water = 10 $cm^{3}$

Put the value of density of acetone and its mass in equation (1)

0.7857 $\frac{g}{cm^{3}}$ = $\frac{10 g}{volume}$

Volume of acetone = 12.72 $cm^{3}$

Thus, volume of acetone is more than volume of water because the density of acetone is lower.

6 0

3 years ago

Robert has pure samples of both D-ribose and D-arabinose, but he forgot to label them. He only has some nitric acid, the reagent

Sauron [17]

Answer:

Following are the responses to the given question:

Explanation:

Since HN03 is an oxidation substance D-ribose u.ith oxidized to form in rubric acid Ribose is chiral, but rubric acid is achiral because of its symmetry mirror level, Hence no infrared roster in the sample holder is observed.

Please find the attached file.

D-Arabinose, on either hand, gives optical aldaric acid with such a net optical rotation observed inside the polarimeter for diagnosis with HN03.

4 0

3 years ago