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Anestetic [448]
3 years ago
12

If gravity ceased to exist on Earth, what would happen to our life supporting atmosphere ?

Chemistry
1 answer:
melamori03 [73]3 years ago
4 0
C is correct. right now gravity holds the atmosphere so if there is no gravity then no atmospher
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How much Sr(OH)2 • 8 H2O (M = 265.76) is needed to prepare 250.0 ML of solution in which [OH]= 0.100M?
SVEN [57.7K]
Use the formula 
first step: 
Use the formula 
molarity= mole/liter
change ml to l
plug in data
 to get .1=mole/.25 or .1M*.25liter
which =.025 moles
then divide .025 moles by two because there are two OH in Sr(OH)2 
then multiply that by 265.76 (the molar mass of water) 
.0125*265.76
which is 3.32grams this is your answer 




6 0
3 years ago
Read 2 more answers
Round off the following measurement to three significant figures. 1.296 g
REY [17]
It would be 1 or 2 because if the number is higher than 5 you need to round up , if its lower than 5 you need to round down.

7 0
4 years ago
Ammonia is produced industrially via this exothermic reaction.
Cloud [144]

Answer:

high pressure of 200-300 atm.

low temp. of between 400-500 degrees celsius:this is for continuous development of ammonia since it decomposes at high temp fathermore the reaction is exothermic

a catalyst to speed up the rate of reaction:i guess it is finely divided iron impregnated in aluminium oxide

platinum can be used as a catalyst but it is easily poisoned

hope it helps

Explanation:

6 0
3 years ago
The combustion of ethane ( C 2 H 6 ) produces carbon dioxide and steam. 2 C 2 H 6 ( g ) + 7 O 2 ( g ) ⟶ 4 CO 2 ( g ) + 6 H 2 O (
dexar [7]

Answer:

10.4 moles of CO2 are produced

Explanation:

take the 5.2 moles of C2H6 and multiply that by the mole ratio of CO2 to C2H6 in the reaction (4/2)

5.2 * (4/2) = 10.4

4 0
3 years ago
Read 2 more answers
The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihyd
kenny6666 [7]

Answer:

The answer is "= 0.078 \ kg \ H_2".

Explanation:

calculating the moles in CH_4 =\frac{PV}{RT}

                                                =\frac{(0.58 \ atm) \times (923 \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(232^{\circ} C +273)}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(505)K}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (41.4605 \frac{L \cdot atm}{mol})}\\\\= 12.9 \ mol

Eqution:

CH_4 +H_2O \to  3H_2+ CO \ (g)

Calculating the amount of H_2 produced:

= 12.9 \ mol CH_4 \times  \frac{3 \ mol \ H_2 }{1 \ mol \ CH_4}\times \frac{2.016 g H_2}{1 \ mol \ H_2}\\\\= 78 \ g \ H_2 \\\\= 0.078 \ kg \ H_2

So, the amount of dihydrogen produced = 0.078 \frac{kg}{s}

5 0
3 years ago
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