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Tanya [424]
3 years ago
7

What kind of elements are on the left side of the zig-zag line on the periodic table?

Chemistry
1 answer:
Ber [7]3 years ago
3 0

Metals are to the left of the zig-zag, nonmetals are to the right, and metalloids lie on/beside the line.

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NEED HELP NOW!!!!!!!
Afina-wow [57]

Answer:

fission

Explanation:

hope that helps

3 0
2 years ago
Read 2 more answers
A gas starts at a volume of 23 L and a pressure of 1.23 atm. What is the new pressure if you
Anna71 [15]

Answer:

<h2>1.89 atm</h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question we have

V_2 =  \frac{1.23 \times 23}{15}  =  \frac{28.29}{15}  \\  = 1.886

We have the final answer as

<h3>1.89 atm</h3>

Hope this helps you

7 0
2 years ago
Write the Noble Gas Notation for Silicone
ozzi

Answer:

[Ne] 3s2 3p2

Explanation:

Neon (Ne) is the noble gas right before silicon (Si).

Then right after neon is the 3s subshell. It has two electrons and is full.

After 3s comes the 3p subshell, and silicon only has two electrons in the 3p subshell (you can just count the electrons in each subshell on your periodic table).

8 0
3 years ago
A substance that has a uniform composition and is able to be separated by physical means
azamat
The answer is homogeneous mixture. It is a mixture which has uniform composition and properties all throughout. Mixtures can be separated by physical processes. Mixtures are systems that consist of two or more substances which are mixed but not chemically combined.
7 0
3 years ago
A sample of gas occupies a volume of 27 mL at a temperature of 161K. What is the volume of the temperature if raised to 343K?
d1i1m1o1n [39]

Answer: The  volume is 57.52 mL if the temperature if raised to 343K.

Explanation:

Given: V_{1} = 27 mL,       T_{1} = 161 K

V_{2} = ?,        T_{2} = 343 K

According to Charles law, at constant pressure the volume of an ideal gas is directly proportional to temperature.

Formula used is as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{27 mL}{161 K} = \frac{V_{2}}{343 K}\\V_{2} = 57.52 mL

Thus, we can conclude that volume is 57.52 mL if the temperature if raised to 343K.

4 0
3 years ago
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