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iamastudent79
01.08.2020 •
Chemistry
1) If there are 0.375 moles 15 grams of compound, what would its molar mass be?
2) How many moles are there in 90 grams of magnesium oxide?
3) Calculate the molecular mass of ammonia.
4) Calculate the formula mass of sodium hydrogen sulphate.
5) Assuming the relation atomic mass of metal M to be 56, what would the empirical formula of its oxide containing 70.0% of M be?
I want you to answer all of the questions, 5 of them
And don't answer the question just for points if you don't know what it means
And also show workings I won't accept an answer without an explanation
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Ответ:
1) 40
2) 2.25 moles
3) 17
4) 120
5) Fe₂O₃
Explanation:
Please see attached picture for full solution.
Ответ:
1)![\boxed{Molar \ mass = 40\ g/mol}](/tpl/images/0716/4028/abdcc.png)
2)![\boxed{No.\ of\ Moles = 1.61 \ moles}](/tpl/images/0716/4028/71f39.png)
3)![\boxed{Ammonia = 16\ g/mol}](/tpl/images/0716/4028/4aee8.png)
4)![\boxed{Sodium\ Hydrogen\ Sulphate= 120 \ g/mol}](/tpl/images/0716/4028/68b72.png)
5)![\boxed{Fe_{2}O_{3}}](/tpl/images/0716/4028/dbe0a.png)
Explanation:
Question # 1:
Using Formula, No. of moles = Mass in grams / molar mass
Where moles = 0.375 mol. , mass = 15 g
=> Molar Mass = Mass in grams / No. of moles
=> Molar Mass = 15 / 0.375
=> Molar mass = 40 g/mol
Question # 2:
No. of moles = Mass in grams / Molar mass
Where Mass = 90 g ,
=> Molar Mass =![MgO_{2}](/tpl/images/0716/4028/0d9e8.png)
=> MM = 24 + (16*2)
=> MM = 24 + 32
=> MM = 56 g/mol
No. of Moles = 90 / 56
No. of Moles = 1.61 moles
Question # 3:
Ammonia =>![NH_{3}](/tpl/images/0716/4028/5e30f.png)
N has atomic mass 14 ang H has atomic mass 1
Ammonia = (14)+(1*2)
Ammonia = 14+2
Ammonia = 16 g/mol
Question # 4:
Sodium Hydrogen Sulphate =>![NaHSO_{4}](/tpl/images/0716/4028/77311.png)
Where Na has atomic mass 23, H has atomic mass 1 , sulpher 32 and oxygen 16
=> 56+64
=> 120 g/mol
Question # 5:
The metal which has relative atomic mass of 56 is Iron (Fe)
Given that the oxide contains 70.0 % of Metal
Mass of
(Metal oxide) = 56 / 70 * 100
Mass of
(Metal oxide) = 0.8 * 100
Mass of
(Metal oxide) = 80
Mass of x O's = 80 - 56
Mass of x O's = 24
Now, Mass of O = 24 / 16
Mass of O = 1.5
So, Fe Metal and its oxide are in the ratio of
1 : 1.5
×2 ×2
2 : 3
So, the empirical formula of metal with its oxide is![Fe_{2}O_{3}](/tpl/images/0716/4028/91459.png)
Ответ:
I WANT TO TALK ITS SAME HERE BORED TOO LOL