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Vinvika [58]
3 years ago
14

How can molecules with polar bonds be nonpolar?

Chemistry
1 answer:
Alexandra [31]3 years ago
5 0
A molecule can<span> possess </span>polar bonds<span> and still be </span>non polar. If the polar bonds<span> are evenly (or symmetrically) distributed, the </span>bond<span> dipoles cancel and </span>do<span> not create a molecular dipole.</span>
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a student used 10 mL water instead of 30 mL for extraction of salt from mixture. How may this change the percentage of NaCl extr
Anit [1.1K]
It will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.

If it is known that solubility of NaCl is 360 g/L, let's find out how many NaCl is in 30 mL of water:

360 g : 1 L = x g : 30 mL

Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 30 mL

Now, crossing the products:
x </span>· 1,000 mL = 360 g · 30 mL
x · 1,000 mL = 10,800 g mL
x = 10,800 g ÷ 1,000 
x = 10.8 g

So, from 30 mL mixture, 10.8 g of NaCl could be extracted.

Let's calculate the same for 10 mL water instead of 30 mL.

360 g : 1 L = x g : 10 mL

Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 10 mL

Now, crossing the products:
x </span>· 1,000 mL = 360 g · 10 mL
x · 1,000 mL = 3,600 g mL
x = 3,600 g ÷ 1,000 
<span>x = 3.6 g
</span>
<span>So, from 10 mL mixture, 3.6 g of NaCl could be extracted.
</span>
Now, let's compare:
If from 30 mL mixture, 10.8 g of NaCl could be extracted and <span>from 10 mL mixture, 3.6 g of NaCl could be extracted, the ratio is:
</span>3.6/10.8 = 1/3

Therefore, i<span>t will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water. </span>
5 0
3 years ago
A student wants to test if giving dogs a toy to play with makes their tail wag more. He gives dog toys to one group, while anoth
kicyunya [14]

Answer:

Experimental: the dogs who get the toys

Control: the dogs who dont recieve a toy

Explanation:

expermental is the group that recieved the tested variable in this case that being the toy

control is the group that doesn't recieve the tested variable

4 0
3 years ago
An engineer is trying to create a fast reaction in a fuel mixture. Which should she do? A. Decrease the amount of fuel B. Lower
oksano4ka [1.4K]
To increase the rate of a reaction, you can either do any of the following:
-increase the temperature
-increase concentration of the aqueous reactant
-increase pressure of the gas
-use a catalyst
-increase surface area of the solid reactant

From these, the engineer should therefore do C. Adding a catalyst speeds up the reaction without really joining the reaction.
3 0
4 years ago
As a reaction proceeds, the ratio between the rate of consumption of reactant and the rate of formation of product:
wariber [46]

Answer:

Depends on the reaction.

Explanation:

Hello,

In this case, the answer is depends on the reaction since the ratios between the rates of both consumption and formation depend upon the stoichiometric coefficients in the chemical reaction. For instance, for the reaction:

A -> 2B

The relationship is:

\frac{1}{-1}r_A =\frac{1}{2} r_B

Therefore, we can see that the rate of consumption of A half the rate of formation of B, but is we consider the following chemical reaction:

2A -> B

The relationship is:

\frac{1}{-2}r_A =\frac{1}{1} r_B

Therefore we can see that the rate of consumption of A doubles the rate of consumption of B.

Best regards.

3 0
3 years ago
Oxygen, Sulfur, and Selenium make up a triad. The atomic mass of oxygen is 15.999 amu, the atomic mass of selenium is 78.96 amu.
3241004551 [841]

Answer:

Explanation:

Selenium is a chemical element with the symbol Se and atomic number 34.

...

Selenium

Melting point 494 K ​(221 °C, ​430 °F)

Boiling point 958 K ​(685 °C, ​1265 °F)

Density (near r.t. ) gray: 4.81 g/cm3 alpha: 4.39 g/cm3 vitreous: 4.28 g/cm3

when liquid (at m.p. ) 3.99 g/cm3

6 0
3 years ago
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