15. Triglycerides, waxes, and steroids are all ___ lipids because they contain only carbon, hydrogen, and oxygen

The very top layer of Earth is the crust, which is relatively quite thin when compared to the other layers of Earth. Which of th

vlabodo [156]

Answer:

D

Explanation:

3 0

3 years ago

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0.41g of neon is held in a 200. mL container at 11 °C. Calculate the pressure in atm.

777dan777 [17]

Answer:

2.39 atm

Explanation:

We'll begin by calculating the number of mole in 0.41 g of neon (Ne). This can be obtained as follow:

Mass of Ne = 0.41 g

Molar mass of Ne = 20 g/mol

Mole of Ne =.?

Mole = mass / molar mass

Mole of Ne = 0.41 / 20

Mole of Ne = 0.0205 mole

Next we shall convert 200 mL to L.

1000 mL = 1 L

Therefore,

200 mL = 200 mL × 1 L / 1000 mL

200 mL = 0.2 L

Next, we shall convert 11 °C to Kelvin temperature.

T(K) = T(°C) + 273

T(°C) = 11 °C

T(K) = 11 + 273

T (K) = 284 K

Finally, we shall determine the pressure. This can be obtained as follow:

Mole of Ne (n) = 0.0205 mole

Volume (V) = 0.2 L

Temperature (T) = 284 K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure (P) =?

PV = nRT

P × 0.2 = 0.0205 × 0.0821 × 284

P × 0.2 = 0.4779862

Divide both side by 0.2

P = 0.4779862 / 0.2

P = 2.49 atm

Therefore, the pressure of the gas is 2.39 atm

3 0

3 years ago

In water, Vanillin, C8H8O3, has a solubility of 0.070 moles of vanillin per liter of solution at 25C. What will be produced if 5

Rufina [12.5K]

Answer:

The full amount (5.00 g) will be dissolved in 1 L of water at 25°C.

Explanation:

The molecular weight (MW) of Vanillin (C₈H₈O₃) is calculated from the chemical formula as follows:

MW(C₈H₈O₃) = (12 g/mol x 8) + (1 g/mol x 8) + (16 g/mol x 3) = 152 g/mol

If 0.070 mol of C₈H₈O₃ are soluble per liter of water at 25°C, the maximum mass that can be dissolved in 1 L is:

0.070 mol x 152 g/mol = 10.64 g

Since 5.00 g is lesser than the maximum amount that can be dissolved (10.64 g), the added amount will be completely dissolved in 1 L of water at 25°C.

7 0

3 years ago

alculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer. Express your answer using two decima

svetlana [45]

A 1.0-L buffer solution contains 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.

Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

Answer:

The pH of this solution = 5.06

Explanation:

Given that:

number of moles of CH3COOH = 0.100 mol

volume of the buffer solution = 1.0 L

number of moles of NaC2H3O2 = 0.100 mol

The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

we know that concentration in mole = Molarity/volume

Then concentration of [CH3COOH] = $\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}$ = 0.10 M

The chemical equation for this reaction is :

$\mathtt{CH_3COOH + OH^- \to CH_3COO^- + H_2O}$

The conjugate base is CH3COO⁻

The concentration of the conjugate base [CH3COO⁻] is = $\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}$

= 0.10 M

where the pka (acid dissociation constant)for CH3COOH = 4.74

If 0.035 mol of NaOH is added to the original buffer, the concentration of NaOH added will be = $\mathtt{ \dfrac{0.035 \ mol}{ 1.0 \ L }}$ = 0.035 M

The ICE Table for the above reaction can be constructed as follows:

$\mathtt{CH_3COOH \ \ \ + \ \ \ \ OH^- \ \ \to \ \ CH_3COO^- \ \ \ + \ \ \ H_2O}$

Initial 0.10 0.035 0.10 -

Change -0.035 -0.035 + 0.035 -

Equilibrium 0.065 0 0.135 -

By using Henderson-Hasselbalch equation:

The pH of this solution = pKa + log $\mathtt{\dfrac{CH_3COO^-}{CH_3COOH}}$

The pH of this solution = 4.74 + log $\mathtt{\dfrac{0.135}{0.065}}$

The pH of this solution = 4.74 + log (2.076923077 )

The pH of this solution = 4.74 + 0.3174

The pH of this solution = 5.0574

The pH of this solution = 5.06 to two decimal places

7 0

3 years ago

Estradiol is a female sexual hormone that causes maturation and maintenance of the female reproductive system. Elemental analysi

Ipatiy [6.2K]

Answer:

The molecular formula of estradiol is: $C_{18}H_{24}O_2$ .

Explanation:

Molar mass of of estradiol = M= 272.37 g/mol

Let the molecular formula of estradiol be $C_xH_yO_z$

Percentage of an element in a compound:

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

Percentage of carbon in estradiol :

$\frac{x\times 12 g/mol}{272.37 g/mol}\times 100=79.37\%$

x = 18.0

Percentage of hydrogen in estradiol :

$\frac{y\times 1g/mol}{272.37 g/mol}\times 100=8.88\%$

y = 24.2 ≈ 24

Percentage of oxygen in estradiol :

$\frac{z\times 16 g/mol}{272.37 g/mol}\times 100=11.75\%$

z = 2

The molecular formula of estradiol is: $C_xH_yO_z:C_{18}H_{24}O_2$

7 0

3 years ago