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3 years ago

A colloidal liquid suspended in a gas is called an aerosol. true or false

Chemistry

2 answers:

iragen [17]3 years ago

6 0

Aerosol:

Aerosol is a colloidal dispersion of liquid or solid in gas.

Thus a colloidal liquid suspended in a gas is called an aero sol is true

Aerosol is an example of colloidal dispersion and when a beam of light is fall on colloids then light is scattered means light is spread in all direction. It is a characteristic property of collides.

ella [17]3 years ago

6 0

I think the correct answer is true. A colloidal liquid suspended in a gas is called an aerosol. It can be directly emitted into the atmosphere or they can be formed in the atmosphere by condensation. Hope this answers the question. Have a nice day.

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Ammonia is produced by the Haber process. The equation is shown.

choli [55]

Answer:

option D

Explanation:

Increasing the temperature increases the yield of ammonia and speeds up the reaction as chemical reaction is affected by temperature.

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2 years ago

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____ HBr + ____ Mg(OH)2 ---> ____ MgBr2 + ____ H2O

swat32

Hello Camkirkland,
I think that you are trying to balance this equation.
In order to balance a chemical equation, the numbers of atoms of each element must be equal on both sides of the equation.

In this particular equation, the answer would be (2) HBr + (1) Mg(OH)2 ---> (1) MgBr2 + (2) H2O.

Hope this answers your question!

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3 years ago

Question 1<br> 1 pts<br> How many mols of bromine are present in 35.7g of<br> Tin(IV) bromate?

sleet_krkn [62]

Answer:

n = 0.0814 mol

Explanation:

Given mass, m = 35.7g

The molar mass of Tin(IV) bromate, M = 438.33 g/mol

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No. of moles = given mass/molar mass

So,

$n=\dfrac{35.7}{438.33}\\\\n=0.0814\ mol$

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3 years ago

For each of the esters provided, identify the alcohol and the carboxylic acid that reacted.

Veronika [31]

Answer:

52. The alcohol USED => methanol, CH3OH

The carboxylic acid USED => propanoic acid, CH3CH2COOH.

53. The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

Explanation:

52. To obtain Methyl propanoate, CH3CH2COOCH3, we simply react propanoic, CH3CH2COOH and methanol, CH3OH together as shown below:

CH3CH2COOH + CH3OH —> CH3CH2COOCH3 + H2O

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The carboxylic acid used: propanoic acid, CH3CH2COOH.

53. To obtain Ethyl methanoate, HCOOCH2CH3, we simply react

Formic acid, HCOOH and ethanol, CH3CH2OH together as show below:

HCOOH + CH3CH2OH —> HCOOCH2CH3 + H2O

The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

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3 years ago

Question 8 (10 points)

I am Lyosha [343]

Answer:

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3 years ago

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