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rodikova [14]
3 years ago
10

How do elements on the right side of the periodic table differ from elements on the left side of the table?

Chemistry
1 answer:
lozanna [386]3 years ago
4 0

The answer is c.


Elements on the left side of the table are metals, such as sodium, lithium, potassium, etc.


Elements on the right side are non metals, such as Chlorine, Fluorine, Bromine, etc.

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lilavasa [31]

Answer:

first question

A) it is not testable

second question

C) chemistry

7 0
3 years ago
Can some one work out this equation for me :)
Law Incorporation [45]

Yo sup??

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4 0
2 years ago
What volume of 0.305 m agno3 is required to react exactly with 155.0 ml of 0.274 m na2so4 solution? hint: you will want to write
LuckyWell [14K]

The balanced chemical equation for reaction of AgNO_{3} and Na_{2}SO_{4} is as follows:

2 AgNO_{3}+Na_{2}SO_{4}\rightarrow 2NaNO_{3}+Ag_{2}SO_{4}

From the balanced chemical equation, 2 mol of AgNO_{3} reacts with 1 mol of  NaNO_{3}.

First calculating number of moles of NaNO_{3} as follows:

M=\frac{n}{V}

On rearranging,

n=M\times V

Here, M is molarity and V is volume. The molarity of NaNO_{3}  is given 0.274 M or mol/L and volume 155 mL, putting the values,

n=0.274 mol/L\times 155\times 10^{-3}mL=0.04247 mol

Since, 1 mol of NaNO_{3}  reacts with 2 mol of  AgNO_{3} thus, number of moles of  AgNO_{3}  will be 2\times 0.04247 mol=0.08494 mol.

Now, molarity of  AgNO_{3} is given 0.305 M or mol/L thus, volume can be calculated as follows:

V=\frac{n}{M}=\frac{0.08494 mol}{0.305 mol/L}=0.2785 L=278.5 mL

Therefore, volume of  AgNO_{3} is 278.5 mL.

4 0
3 years ago
6th grade science i mark as brainliest​
Blababa [14]

Answer:

It's the 4th one. The car is accelerating gradually.

8 0
3 years ago
HELPP 20+ POINTS!!!
VashaNatasha [74]
The answer is •c•
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2 years ago
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