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Anettt [7]
3 years ago
6

Name 2 parts of a microscope

Chemistry
1 answer:
Airida [17]3 years ago
6 0
Two parts are stage and coarse focus
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What is magnesium used for in its pure form
Sloan [31]

Answer: The pure metal has low structural strength, magnesium is mainly  used in the form of alloys- principally with 10% or less of aluminum, zinc, and manganese- to improve it's hardness, tensile strength, and ability to be cast, welded, and machined.

I hope this really helps you!!! :) And Have a fantastic day!!!

 

6 0
3 years ago
Is there are 10.0 g of sucrose and 8.0 g of oxygen, how many moles of sucrose are available for this reaction
liq [111]
If there is a multiple choice then the answer would be 0.029. I hope this helps you
8 0
3 years ago
Read 2 more answers
Rank the following bonds in order of decreasing bond length, based on periodic trends, starting with the longest bond at the top
sattari [20]

Answer:

H-BI,H-Se,H-S,H-I,H-Br

Explanation:

One thing that must be kept in mind is that atomic size increases down the group and decreases across the period. The bond lengths of species are influenced by the relative sizes of atoms or ions present in the bond.

The bonds in the answer have been arranged on basis of their decreasing atomic size because the greater the atomic size of the atoms, the greater the bond length and vice versa.

5 0
3 years ago
Round to 4 significant figures.<br> 35.5450
Lelu [443]
Answer:
35.5450 will be rounded to 35.55
Explanation:
=35.5450
if the last digit is less than 5 then it will be ignored
=35.545
when the dropping digit is 5 then the retaining digit will increse by a factor of 1
=35.55
i hope this will help you
3 0
3 years ago
Read 2 more answers
Calculate the following quantity: molarity of a solution prepared by diluting 45.45 mL of 0.0404 M ammonium sulfate to 550.00 mL
dybincka [34]

Answer:

M_2=3.34x10^{-3}M

Explanation:

Hello!

In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

M_1V_1=M_2V_2

Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:

M_2=\frac{M_2V_2}{V_1} =\frac{45.45mL*0.0404M}{550.00mL}\\\\M_2=3.34x10^{-3}M

Best regards!

5 0
2 years ago
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