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pav-90 [236]
3 years ago
15

Object A has a mass of 500g and object B has a mass of 300g. Which object is more dense?

Chemistry
1 answer:
KonstantinChe [14]3 years ago
4 0

Answer:

500g because 300g is less dense than 500g its kinda like a weight

Explanation:

i

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Which statement explains the method to adjust the strength of the flame on the Bunsen burner?
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Answer:

adjusting the air mix by rotating the barrel and adjusting the gas with the needle valve to obtain a flame of suitable height and intensity

4 0
4 years ago
Lithium (Li) and oxygen (O) are both in period 2. They both have _____.
AlladinOne [14]

They both have two electron shells

<h3>Further explanation</h3>

The period 2 element lies in the second row of the periodic system.

Consists of the elements: lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, and neon

  • Lithium (Li)

atomic number : 3

electron configuration : [He] 2s¹

atomic number = number of proton=number of electron(in neutral atom)

So Li have 3 protons and 3 electrons

Because it fills the 2s orbital it has 2 shells

  • Oxygen (O)

atomic number : 8

electron configuration :  [He] 2s²2p⁴

So O have 8 protons and 8 electrons

Because it fills the 2s and 2p orbital it has 2 shells

So Lithium (Li) and Oxygen (O) are both have two electron shells

6 0
3 years ago
If the kitchen door is closed, how will that affect your ability to smell cooking odors in your room?
Aleonysh [2.5K]

Answer:Even if your door is closed, you would still smell the odors because of the space under the door and the space that is needed to close the door.  

Explanation:

6 0
3 years ago
Read 2 more answers
A 52.0 mL portion of a 1.20 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a
RoseWind [281]

Answer:

C_3=0.125M

Explanation:

Hello!

In this case, we can divide the problem in two steps:

1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

V_1C_1=V_2C_2

So we solve for C2:

C_2=\frac{C_1V_1}{V_2}=\frac{52.0mL*1.20M}{278mL}\\\\C_2=0.224M

2. Now, since 111 mL of water is added, we compute the final volume, V3:

V_3=139+111=250mL

So, the final concentration of the 139 mL portion is:

C_3=\frac{139 mL*0.224M}{250mL}\\\\C_3=0.125M

Best regards!

8 0
3 years ago
A 47.0 mL aliquot of a 0.400 M stock solution must be diluted to 0.100 M. Assuming the volumes are additive, how much water shou
mel-nik [20]

Answer:

The answer is 0.188L

Explanation:

I did the problem on paper and put the answer on the pictures. I'm sorry if I didn't explain it well but I hope I helped.

5 0
3 years ago
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