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9 days ago

A solution of water (kb=0.512 ?c/m) and glucose boils at 103.06 ?c. what is the molal concentration of glucose in this solution?

assume that the boiling point of pure water is 100.00 ?c.

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In NMR spectroscopy, the strong magnetic field establishes an energy gap between the alpha and beta spin states, which enables t

Anarel [2989]

Answer:

In the context of NMR spectroscopy, a significant magnetic field creates an energy difference between the alpha and beta spin states, which allows nuclei to absorb RF radiation, ultimately leading to the excitation of a nucleus from a +1/2 spin state to a -1/2 spin state.

Explanation:

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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:

= $C_xH_yO_zN_w= C_9H_{13}O_3N_1$

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3 months ago

Which of the following is a reasonable ground-state electron configuration?

lorasvet [2795]

Answer:

The correct choice is: option A.

Justification:

To address this inquiry, we need to evaluate the total number of electrons each orbital can accommodate.

Orbital Number of electrons

s 2

p 6

d 10

f 14

Provided options:

A. 1s² 2s² 2p⁶ 3s² This configuration is valid as it aligns with the permitted number of electrons in each orbital and follows the correct sequence.

B. 1s² 2s² 2p⁶ 3s² 3d⁴ This configuration is not accurate because

3d⁴ should follow 3p.

C. 1s² 2s² 2d¹⁰ 2p³ This is incorrect since 2d¹⁰ is not a valid orbital.

D. 1s² 2s^s 2p³ 2d¹⁰ This option contains two errors; s as an exponent does not exist, and 2d¹⁰ is also an invalid description.

3 0

2 months ago

Of elements N, O, Cl, Na, and Which two would likely have similar chemical properties and why

lions [2927]

Oxygen and Nitrogen are the most alike among the listed elements due to their proximity in the periodic table. While this reasoning may not be particularly robust, these two elements share certain similarities. They are both classified as non-metals, exhibit high electronegativity, exist as diatomic gases in their elemental forms, contain a similar number of valence electrons, and generally act as oxidizing agents. Although Oxygen and Chlorine also show similarities, they are not as closely related as Oxygen and Nitrogen.

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2 months ago