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xz_007 [3.2K]
3 years ago
6

How do the atmosphere conditions near the beginning of Precambrian time contrast with the atmosphere conditions that are present

now?
Chemistry
1 answer:
babunello [35]3 years ago
5 0
The early precambrian atmosphere consisted primarily of nitrogen and carbon dioxide with almost no oxygen. 

<span>Today, the atmosphere contains about 20% oxygen, less carbon dioxide and similar amounts of nitrogen. </span>

<span>Photosynthetic green-leaf plants and trees are largely responsible for the change, converting carbon dioxide to oxygen.</span>
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How many moles are in 187.54 grams of magnesium chlorate?
Svetlanka [38]

Hey there!

Magnesium chlorate: Mg(ClO₃)₂

Find molar mass.

Mg: 1 x 24.305 = 24.305

Cl: 2 x 35.453 = 70.906

O: 6 x 16 = 96

------------------------------------

                      191.211 g/mol

We have 187.54 grams.

187.54 ÷ 191.211 = 0.9808

There are 0.9808 moles in 187.54 grams of magnesium chlorate.

Hope this helps!

3 0
3 years ago
How many milliliters of a 5.0 M H2SO4 stock solution would you need to prepare 108.0 mL of 0.45 M H2SO4?
PIT_PIT [208]
For the purpose we will use solution dilution equation:
c1xV1=c2xV2
Where, c1 - concentration of stock solution; V1 - a volume of stock solution needed to make the new solution; c2 - final concentration of new solution; V2 - final volume of new solution.
c1 = 5.00 M
c2 = 0.45 M
V1 = ?
V2 = 108 L
When we plug values into the equation, we get following:
5 x V1 = 0.45 x 108
<span>V1 = </span>9.72 L
7 0
3 years ago
Question 2: Which of the following is NOT true about streams?
mafiozo [28]

Answer:

q 3 c q 2 d

Explanation:

8 0
3 years ago
Lead (II)oxide+calcium metal=​
AfilCa [17]
I didn’t understand
3 0
3 years ago
8. When a 2.5 mol of sugar (C12H22O11) are added to a certain amount of water the boiling point is raised by 1 Celsius degree. I
Julli [10]

Hey there!:

8) ΔTb = i*Kb*m  

 m is molality

 Since same number of mol is added to same amount of water in both cases

m will be same for both

is 1 for glucose since it is covalent compound

is 4 of Al(NO3)3 as it breaks into 1 Al₃⁺ and 3 NO₃⁻

So,  ΔTb will be 4 times in aluminum nitrate case

So, boiling point will change by 4ºC


9) use Q = m*  L

L =  heat of vaporization so:

T1=T2=100ºC

5.40 * 1000 => 5400  cal/g

Q =   5400 / 540

Q = 10 grams


Hope that thlps!

5 0
3 years ago
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