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frez [133]
4 years ago
7

If 10.0g of Al2(SO3) is reacted with 10.0g of NaOH, determine the limiting reagent

Chemistry
1 answer:
ehidna [41]4 years ago
3 0

(10.0 g Al2(SO3)3) / (294.1544 g Al2(SO3)3/mol) = 0.033996 mol Al2(SO3)3 (10.0 g NaOH) / (39.99715 g NaOH/mol) = 0.25002 mol NaOH 
0.033996 mole of Al2(SO3)3 would react completely with 0.033996 x (6/1) = 0.203976 mole of NaOH, but there is more NaOH present than that, so NaOH is in excess and Al2(SO3)3 is the limiting reactant. 
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I need help like extremely bad
Anna007 [38]

Answer:

First start with the ones we know

Explanation:

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a pair - so must be bigger than one chromosome

1. small - gene

2.chromosome - chromosomes contain genes so they must be bigger

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now 5.

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7 0
3 years ago
Which of the following occupies a volume? solids liquids gases all of the above
yanalaym [24]
Solids, liquids and gases all take up volume. 
4 0
3 years ago
Read 2 more answers
Balance the following equations:
yarga [219]

Explanation:

C3H8 + 502 ----> 3CO2 + 4H20

4NH3+7O2------> 4NO2+6H2O

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4 0
3 years ago
You find a clean 100-ml beaker, label it "#1", and place it on a tared electronic balance. You add small amount of unknown solid
Ilia_Sergeevich [38]

Answer:

the mas is .291 g

Explanation:

the mass of a object does not change. so when added the substance the beaker. you had the mass of both objects together. you know the mass of the beaker and you know the mass of both. since mass does not change. the beakers mass is still 74.605g. the mass of both objects is 74.896. all you have to do is subtract the mass of the beaker from the total mass. 74.896 - 74.605 equals .291g. so the mass of the unknown substance Is .291g

7 0
3 years ago
5K + KNO3 ------ 3K2O + N
masya89 [10]

Answer:

174,957.143 grams of potassium and 89,228.478 grams of potassium nitrate will be needed.

Explanation:

5K +KNO_3\rightarrow 3K_2O + N

Mass of nitrogen =  27 lbs = 12,247 g

1 lbs = 453.592 g

Moles of nitrogen = \frac{12,247 g}{14 g/mol}=874.786 mol

According to reaction, 1 mole of nitrogen is produced from 5 moles of potassium and 1 mole of potassium nitrate.

Then 874.786 mol of nitrogen will be obtained from :

\frac{5}{1}\times 874.786 mol=4,373.928 mol of potassium.

Then 874.786 mol of nitrogen will be obtained from :

\frac{1}{1}\times 874.786 mol=874.789 mol of potassium nitrate.

Mass of 4,373.928 moles of potassium:

4,373.928 mol\times 40 g/mol=174,957.143 g of potassium

Mass of 874.789 moles of potassium nitrate:

874.789 mol\times 102 g/mol=89,228.478 g of potassium nitrate

3 0
3 years ago
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