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Tom [10]
3 years ago
8

How many particles are in 6.02 moles of methane?

Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
5 0
One mole of methane has a mass of 16 g

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Which has a greatest mass, one atom of carbon, one atom of hydrogen, or one atom of litium
zysi [14]

Answer:

One atom of carbon

Explanation:

5 0
3 years ago
Approximately how many atoms make up 2.3 moles of gallium?
Len [333]
I think its b I think
5 0
3 years ago
25 points! Help please
Sedbober [7]

Answer:

I think its b

Explanation:

but I wouldn't depend on this answer

4 0
3 years ago
Read 2 more answers
a physical property can be observed while the substance remains the same in its chemical composition. true false
sweet [91]

While the material retains its chemical makeup, the physical property may be examined. The given statement is true.

The matter can undergo variations in physical or chemical properties. The physical changes of a matter occur when the matter undergoes changes in its physical properties like changes in the state of matter, weight, color, etc.

But the chemical composition of matter will remain constant if it undergoes a physical change. Whereas in chemical change, the matter undergoes a change in the composition of the substance but there will be no change in the physical properties.

Hence, The assertion is correct in that physical properties can be seen while the substance's chemical makeup stays constant.

To learn more about physical and chemical change, visit: brainly.com/question/21509240

#SPJ4

8 0
1 year ago
In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.68 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what
melomori [17]

Answer:

Y=65.7\%

Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

n_{MgI_2}^{theoretical}=2.68molMg*\frac{1molMgI_2}{1molMg} =2.68molMgI_2

Thus, we compute the percent yield as:

Y=\frac{n_{MgI_2}^{real}}{n_{MgI_2}^{theoretical}} *100\%=\frac{1.76mol}{2.68mol} *100\%\\\\Y=65.7\%

Best regards.

8 0
3 years ago
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