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2 years ago

14

Clouds are colloids. what are the dispersed particles

2 answers:

n200080 [17]2 years ago

6 0

The particles in a colloid are large enough to scatter light, a phenomenon called the Tyndall effect. Clouds are colloidal mixtures. They are composed of water droplets that are much larger than molecules, but that are small enough that they do not settle out.

scZoUnD [109]2 years ago

3 0

Answer:

Water droplets slowly forming until they reach a size that can no longer be suspended in the air.

Explanation:

When air becomes saturated with whater, it can no longer maintain water as a gas. Some water molecule come together and stay that way. They, in turn, coallesce with other nearby particles. This continues until the droplets are large enough that they begin to noticeably refract the sunlight, producting the visual effect we call clouds. The droplets are suspended in air - an heterogenous suspension called a colloid. At some point the colloid cloud begins to shed the larger water droplets, usually directly over my spot on the trail.

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Container A (with volume 1.23 L) contains a gas under 3.24 atm of pressure. Container B (with volume

kolbaska11 [484]

Now that a lot of pressure

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3 years ago

QUESTION 5<br> Match the symbol with the type of symbiotic relationship it represents

marshall27 [118]

Answer:

A : 1

B : 3

C : 4

D : 2

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3 years ago

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: Citric acid, H3C6H5O7, is a triprotic acid. Consider a buffer system comprising H2C6H5O7 - and HC6H5O7 2- ions. What is the ne

olchik [2.2K]

Answer:

The Net reaction is

$H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction$
$H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH$
$HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH$

Explanation:

From the Question we are told that the buffers are

$H_2C_6H_5O_7^ -$ and $HC_6H_5O_7 ^{ 2-}$

When NaOH is added the Net ionic reaction would be

$H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction$
$H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH$
$HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH$

8 0

3 years ago

Read 2 more answers

What is the pH of a 3.40 mM of acetic acid, 500 mL in which 50 mL of 1.00 M NaOAc has been added? Ka HOAc is 1.8 x 10-5? What pe

Deffense [45]

Answer:

a) pH = 4.213

b) % dis = 2 %

Explanation:

Ch3COONa → CH3COO- + Na+

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]

mass balance:

⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]

<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)

∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL

⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water

⇒ [ CH3COO- ] = [ H3O+ ] + 1.00

⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5

⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0

⇒ [ H3O+ ] = 6.12 E-5 M

⇒ pH = - Log [ H3O+ ] = 4.213

b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4

∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol

⇒ % dis = 3.4 / 1.7 = 2 %

7 0

3 years ago

A light bulb has a resistance of 96.8 Ω. What current flows through the bulb when it is connected to a 120 V source of electrica

Finger [1]

Answer:

I = 1.23 A

Explanation:

Given that,

The resistance of the lightbulb, R = 96.8 Ω

Voltage, V = 120 V

We need to find the current flows through the lightbulb. Let the current be I. We can use the ohm's law to find it i.e.

$V=IR\\\\I=\dfrac{V}{R}\\\\I=\dfrac{120}{96.8}\\\\I=1.23\ A$

So, the current flows through the bulb is 1.23 A.

8 0

2 years ago