Which sources would preserve the soft tissues of an early mammal? Check all that apply.

In an experiment to determine how to make sulfur trioxide, a chemist combines 32.0 g of sulfur with 50.0 g of oxygen. She finds

LuckyWell [14K]

Answer:

The chemist needs to react 40 g of sulfur with 60 g of oxygen to make 100 g of sulfur trioxide.

Explanation:

2S (s) + 3O₂ (g) → 2SO₃ (g)

64g + 96g → 160 g

32g + 48g → 80 g

x + y → 100 g

1 mol SO₃ ___ 80g

n _______ 100g

n = 1.25 mol SO₃

1 mol S ___ 32 g

1,25 mol S __ 40 g

1 mol O₂ ___ 32 g

1,875 mol O₂ ___ 60 g

4 0

3 years ago

How many of the following are found in 15.0 kmol of xylene (C8H10)? (a) kg C8H10; (b) mol C8H10; (c) lb-mole C8H10; (d) mol (g-a

77julia77 [94]

Answer:

a) 1592.4 kg C8H10

b) 15*10³ mol C8H10

c) 33.1 lb-mole

d) 1.2 *10^5 mol C

e) 1.5 * 10^5 mol H

f) 1.44 * 10^6 grams C

g) 1.52 * 10^5 grams H

h) 9.03*10^27 molecules C8H10

Explanation:

Step 1: Data given

Number of moles C8H10 = 15.0 kmol =15000 moles

Molar mass of C8H10 = 106.16 g/mol

Step 2: Calculate mass C8H10

Mass C8H10 = moles C8H10 * molar mass C8H10

Mass C8H10 = 15000 * 106.16 g/mol

Mass C8H10 = 1592400 grams = 1592.4 kg

Step 3: Calculate moles C8H10

15.0 kmol = 15*10³ mol C8H10

Step 4: Calculate lb-mol

15.0 kmol = 33.1 lb-mole

Step 5: Calculate moles of C

For 1 mol C8H10 we have 8 moles of C

For 15*10³ mol C8H10 we have 8* 15*10³ =1.2 *10^5 mol C

Step 6: Calculate moles H

For 1 mol C8H10 we have 10 moles of H

For 15*10³ mol C8H10 we have 10* 15*10³ = 1.5 * 10^5 mol H

Step 7: Calculate mass of C

Mass C = moles C * molar mass C

Mass C = 1.2 *10^5 mol C * 12 g/mol

Mass C = 1.44 * 10^6 grams

Step 8: Calculate mass of H

Mass H = moles H * molar mass H

Mass H = 1.5 *10^5 mol H * 1.01 g/mol

Mass C = 1.52 * 10^5 grams

Step 9: Calculate molecules of C8H10

Number of molecules = number of moles * number of Avogadro

Number of molecules = 15*10³ mol C8H10 * 6.022*10^23

Number of molecules = 9.03*10^27 molecules

8 0

3 years ago

(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct

Zarrin [17]

Explanation:

The given data is as follows.

$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

$\Lambda^{o}_{m}$ (H-O=C-ONO) = $1.046 \times 10^{-2}$

$\Lambda^{o}_{m}$ (HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

$\Lambda_{m} = \frac{conductivity}{concentration}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

= $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

= $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

= $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

$\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

= 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

= $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

= $0.017973 \times 10^{-2}$

= $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$ .

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

$pK_{a} = -log K_{a}$

= $-log (1.7973 \times 10^{-4})$

= 3.7454

Therefore, value of $pK_{a}$ is 3.7454.

3 0

3 years ago

Which of the following are examples of chemical changes? Select all that apply.

Natali5045456 [20]

Answer:

b. milk spoiling and c. firecrackers exploding

Explanation:

These are both chemical changes, the composition of them change when this happens and it cannot be reversed

6 0

2 years ago

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r

rodikova [14]

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

= 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

$K=Ae^{\frac{-E}{RT}}$

If $K_1=Ae^{\frac{-E_1}{RT}} -------equation 1$ &

$K_2=Ae^{\frac{-E_2}{RT}} -------equation 2$

$\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}$

$E_1= E_2-RT*In(\frac{K_1}{K_2})$

$E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})$

$E_1 = 28957.39292$ $J/mol$

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0

3 years ago