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3 years ago

I'm afraid I won't pass. And very nervous. Can you help me please.

1 answer:

4 0

Just look at the number in front also called coefficient (you have to balance the equations first, but all the questions here are balanced, so no worries). for q1.
in the balanced equation, the number in front of aluminum oxide is 2 (2 - this number Al2O3) and for aluminium is 4 as in (4 Al). so the ratio is 2:4. simplified it is 1:2. or write it out fully
2 Al2O3: 4 Al
ignore everything after the number.
2:4
same as 1:2
Aluminium oxide to oxygen
2 Al2O3: 3 O2
2:3
aluminum to oxygen
4 Al: 3 O2
4:3

question 2
Mercury oxide to Mercury
2 HgO : 2 Hg
2:2
same as 1:1
Mercury oxide to oxygen
2 HgO : O2
since oxygen in this case does not have a number written in front of it, the default is 1.
2: 1.
you should be able to do the rest

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How many atoms are there in 1.3 x 10^22 molecules of N(O)2

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1.3 x 10^22 times 3 = 3.9 x 10^22 atoms

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2 years ago

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What is the SI base unit for time?

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2 years ago

A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

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3 years ago

HELP IM TIMEDDDDDDDDDDDD

nadya68 [22]

Answer:

ethier a dessert or a plains

though plains can get rain in the summer

from me living in both

it seems more like a dessert

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3 years ago

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If 5 moles of P4 reacted with 22 moles Cl2 according to the above reaction, determine:

GaryK [48]

Solution:

$P_4+6Cl_2\rightarrow 4PCl_3$

a) By stoichoimetry If 6 moles of $Cl_2$ gives 4 mole of $PCl_3$ , then 22 moles of $Cl_2$ will give: $\frac{4}{6}\times 22=\frac{44}{3}=14.66$ moles of $PCl_3$

b) If 6 moles of $Cl_2$ reacts with one mole of $P_4$ , then 22 moles of $Cl_2$ will react : $\frac{1}{6}\times 22=\frac{11}{3}=3.66$ moles of $P_4$

Moles of $P_4$ left after the reaction = $5-3.66=\frac{4}{3}=1.34$ moles.

c) 0 Moles of $Cl_2$ , since it is present in less amount it will get completely consumed in the reaction. Hence, it is limiting reagent.

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3 years ago