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sleet_krkn [62]
3 years ago
7

True or false, The universal gas constant, R, is . 0821 when we use mm Hg as our unit for pressure

Chemistry
1 answer:
ludmilkaskok [199]3 years ago
5 0

Answer:

This is true.

Explanation:

R should be 0.0821

Good luck!

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The table shows the volumes of different samples of liquid? Which sample has the greatest DENSITY.
arlik [135]

Answer:

Maybe B

Explanation:

4 0
4 years ago
Read 2 more answers
What is the time required for contamination source
Afina-wow [57]
Hi there!

It can take years to remove all of the harmful substances from the water.

I hope my answer helped :)
8 0
3 years ago
There are two steps in the extraction of copper metal from chalcocite, a copper ore. In the first step, copper(I) sulfide and ox
maxonik [38]

Answer:

2Cu2S + 3O2 + 2C -------> 4Cu + 2SO2 + 2CO

Explanation:

Equation 1 should correctly be written as;

2Cu2S + 3O2-----> 2Cu2O + 2SO2

Equation 2 should be correctly written as;

2Cu2O + 2C -----> 4Cu + 2CO

The overall reaction equation is;

2Cu2S + 3O2 + 2C -------> 4Cu + 2SO2 + 2CO

Note that species that are intermediates are cancelled out .

4 0
3 years ago
How is burning magnesium different than burning methane
olya-2409 [2.1K]

You can stop the burning of methane with water or carbon dioxide extinguishers but problems arise when you try to use this to stop the burning of the magnesium.

Explanation:

To burn magnesium (Mg) and methane (CH₄) you need to react them with oxygen:

2 Mg (s) + O₂ (g) → 2 MgO + heat

CH₄ (g) + 2  O₂ (g) → CO₂ (g) + 2 H₂O (g) + heat

However at that temperatures magnesium (Mg) is able to react with water (H₂O) and carbon dioxide (CO₂).

Mg (s) + 2 H₂O (l) → Mg(OH)₂ (s) + H₂ (g)

2 Mg (s) + CO₂ (g) → 2 MgO (s) + C (s)

So the safe option to stop the burning of the magnesium is to limit the oxygen in the air.

we have used the following notations:

(s) - solid

(g) - gas

(l) - liquid

Learn more about:

combustion reactions

brainly.com/question/13824679

#learnwithBrainly

6 0
3 years ago
4Na + O2 2Na2O
beks73 [17]
4
N
a
+
O
2
→
2
N
a
2
O
.
By the stoichiometry of this reaction if 5 mol natrium react, then 2.5 mol
N
a
2
O
should result.
Explanation:
The molecular mass of natrium oxide is
61.98

g
⋅
m
o
l
−
1
. If
5

m
o
l
natrium react, then
5
2

m
o
l
×
61.98

g
⋅
m
o
l
−
1

=

154.95

g
natrium oxide should result.
So what have I done here? First, I had a balanced chemical equation (this is the important step; is it balanced?). Then I used the stoichiometry to get the molar quantity of product, and converted this molar quantity to mass. If this is not clear, I am willing to have another go
5 0
3 years ago
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