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siniylev [52]
3 years ago
8

An electric iron is used to remove wrinkles from clothes the electric iron is made of metal which property of this substance is

most useful for this application
boiling point

density

electrical conductivity

thermal conductivity
Chemistry
1 answer:
Jobisdone [24]3 years ago
5 0

thermal conductivity

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What is the acceleration due to gravity of the moon<br>​
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Answer:

Earth's average surface gravity is about 9.8 meters per second per second. When an object is tossed off a building top or a cliff apex, for instance, it accelerates toward the ground at 9.8 meters per second per second. The Moon's surface gravity is about 1/6th as powerful or about 1.6 meters per second per second,

Explanation:

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Which of the following is the correct wedge and dash conformation for the following Newman projection?
kakasveta [241]

We have that  the correct wedge and dash conformation for the following Newman projection is

IV

From the Diagrams above

Two CH_3 groups points on opposite sides in plane

Two Br are on same side  of plane

Two H also on same side of the plane so the plausible  structure is IV

Therefore

The Correct option is IV

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3 years ago
HELP!!!! I WILL MARK BRAINLIEST WHOEVER ANSWERS FIRST
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Answer:

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2 years ago
Describe what happens in a condensation reactions
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5 0
3 years ago
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

7 0
2 years ago
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