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padilas [110]
2 years ago
7

When cleaning a buret, begin by coating the inside with:.

Chemistry
1 answer:
vladimir2022 [97]2 years ago
3 0

Answer:

I think the answer is distilled water...

Explanation:

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BRAINLIEST AWARD PLZ HELP :)
Vera_Pavlovna [14]
Properties of metals:
High melting points
High density
Ductile
Malleable
Good conductors of electricity
Good conductors of heat


I think if you added a proton you would have chlorine.


The noble gasses are the he chemical elements in group 18 of the periodic table. The gasses in this family include helium, neon, argon, krypton, xenon, and radon. All these gasses are colorless are oderless, elements in this family have atoms with a full outer shell of electrons. They are also called inert gasses.

Six valence electrons
6 0
3 years ago
Read 2 more answers
Consider four atoms from the second period: lithium, beryllium boron carbon and nitrogen which of these element has the lowest e
bearhunter [10]
Lithium has the lowest. if fluorine is the highest then lithium is the lowest. i hope this helps you out!
5 0
3 years ago
Read 2 more answers
Which metal will more easily lose an electron sodium or potassium?
sergey [27]
<span>The metal that would more easily lose an electron would be potassium. It is more reactive than sodium. Also, looking on the periodic table, </span><span>from top to bottom for groups 1 and 2, reactivity increases. So, it should be potassium. Hope this answers the question. Have a nice day.</span>
7 0
3 years ago
Chemistry is the study of all of the following EXCEPT
ValentinkaMS [17]
B - projectile motion
4 0
3 years ago
How many moles are in 1.05 g of gold (Au)?
Wittaler [7]

Answer:

0.005 mol

Explanation:

Moles is denoted by given mass divided by the molecular mass ,  

Hence ,  

n = w / m

n = moles ,  

w = given mass ,  

m = molecular mass .

From the question ,

w = given mass of Gold = 1.05 g ,

m = molecular mass of Gold = 197 g/mol

<u>Hence , moles can be calculated as -</u>

n = w / m = 1.05 g / 197 g/mol = 0.005 mol

7 0
3 years ago
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