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3 years ago

Why do water molecules and the materials dissolved in water move through the plasma membrane slowly ?

1 answer:

4 0

Nonpolar and small polar molecules can pass through the cell membrane, so they diffuse across it in response to concentration gradients. Carbon dioxide and oxygen are two molecules that undergo this simple diffusion through the membrane. The simple diffusion of water is known as osmosis.

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Explain why the solutions of SbCl3(aq) in our lab also contain HCl. Why would adding a few drops of this solution to 300 mL of w

IRISSAK [1]

Answer:

Here's what I get

Explanation:

SbCl₃ reacts with water to form slightly soluble antimony oxychloride.

SbCl₃(aq) +H₂O(ℓ) ⇌ SbOCl(s) + 2HCl(aq)

Your observation is an example of Le Châtelier's Principle in action,

The SbCl₃(aq) in your lab has enough HCl added to push the position of equilibrium to the left and keep the SbOCl in solution.

If a few drops of the SbCl₃(aq) were added to 300 mL of water, the solution would turn cloudy. The HCl would be so dilute that the position of equilibrium would lie to the right, and a cloudy precipitate of antimony oxychloride would form.

7 0

3 years ago

the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal

Sunny_sXe [5.5K]

Rydberg formula is given by:

$\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )$

where, $R_{H}$ = Rydberg constant = $1.0973731568508 \times 10^{7} per metre$

$\lambda$ = wavelength

$n_{1}$ and $n_{2}$ are the level of transitions.

Now, for $n_{1}$ = 2 and $n_{2}$ = 6

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )$

= $1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0278 )$

= $1.0973731568508 \times 10^{7} \times 0.23$

= $0.2523958\times 10^{7}$

$\lambda = \frac{1}{0.2523958\times 10^{7}}$

= $3.9620\times 10^{-7} m$

= $396.20\times 10^{-9} m$

= $396.20 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 5

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.04 )$

= $1.0973731568508 \times 10^{7} \times (0.21 )$

= $0.230 \times 10^{7}$

$\lambda= \frac{1}{0.230 \times 10^{7}}$

= $4.3478 \times 10^{-7} m$

= $434.78\times 10^{-9} m$

= $434.78 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 4

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0625 )$

= $1.0973731568508 \times 10^{7} \times (0.1875 )$

= $0.20575 \times 10^{7}$

$\lambda= \frac{1}{0.20575 \times 10^{7}}$

= $4.8602 \times 10^{-7} m$

= $486.02 \times 10^{-9} m$

= $486.02 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 3

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.12 )$

= $1.0973731568508 \times 10^{7} \times (0.13 )$

= $0.1426585\times 10^{7}$

$\lambda= \frac{1}{0.1426585\times 10^{7}}$

= $7.0097 \times 10^{-7} m$

= $700.97 \times 10^{-9} m$

= $700.97 nm$

5 0

3 years ago

Read 2 more answers

ethylene glycol, commonly used in automobile antifreeze, contains only carbon, hydrogen and oxygen.combustion analysis of a 23.4

Kay [80]

Answer:

Molecular formula of ethylene glycol is C₂H₆O₂.

Explanation:

Given data:

mass of ethylene glycol = 23.4 mg

mass of water = 20.42 mg

mass of carbon dioxide = 33.27 mg

molar mass of ethylene glycol = 62.0 g / mol

Molecular formula of ethylene glycol = ?

Solution:

percentage of carbon = (33.27 mg/ 23.46 mg) × (12 /44) × 100

= (1.4182× 0.27) × 100 = 38.29

percentage of hydrogen = (20.42 mg/ 23.46 mg) × (2 /18) × 100

= (0.8704× 0.11) × 100 = 9.67

percentage of oxygen = 100 - (38.29 + 9.67 )

= 100 - 47.96 = 52.04

Now we will determine the number of grams atoms of carbon, hydrogen and oxygen.

No. of gram atoms of carbon = 38.29 /12 = 3.19

No. of gram atoms of hydrogen = 9.67 / 1 = 9.67

No. of gram atoms of oxygen = 52.04 / 16 = 3.25

Atomic ratio:

C :H :O 3.19/ 3.19 : 9.67 / 3.19 : 3.25 /3.19

C :H :O 1 : 3 : 1

Molecular formula:

Molecular formula = n × (empirical formula)

n = molar mass of compound / empirical formula mass

empirical formula mass= 1 × 12 + 3 × 1 + 1 × 16

empirical formula mass= 12+3 +16 = 31

n = 62 g /mol / 31 = 2

Molecular formula = n × (empirical formula)

Molecular formula = 2 × (CH₃O)

Molecular formula = C₂H₆O₂

6 0

3 years ago

Calculate the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxid

Zielflug [23.3K]

Check the attached file for the answer.

5 0

3 years ago

What type of reaction occurs in a hand warmer?<br> Exothermic<br> Endothermic

storchak [24]

Answer:

The reaction is exothermic.

Explanation:

8 0

2 years ago