Answer:
Explanation:
9.708 mg of CO₂ will contain 12 x 9.708 / 44 = 2.64 g of C .
3.969 mg of H₂O will contain 2 x 3.969 / 18 = .441 g of H .
mg of O in given liquid B = 3.795 - ( 2.64 + .441 ) = .714 mg
ratio of mg of C , H , O in the compound = 2.64 : .441 : .714
ratio of no of atoms of C , H , O in the compound
= 2.64 / 12 : .441 /1 : .714 / 16
= .22 : .44 : .0446
= .22 / .22 : .44 / .22 : .044 / .22
= 1 : 2 : .2
1 x 5 : 2 x 5 : .2 x 5
= 5 : 10 : 1
empirical formula of the compound = C₅H₁₀O
Volume of 89.8 mL at 1 .00 atm at 200⁰C
volume of gas at 1 atm and 0⁰C = 89.8 x 273 / 473 mL
= 51.83 mL
51.83 mL weighs .205 g
22400 mL will weigh .205 x 22400 / 51.83 g
= 88.6 g
So molecular weight = 88.6
Let molecular formula be (C₅H₁₀O)ₙ
molecular weight = n ( 5 x 12 + 10 + 16 )
= 86 n
86 n = 88.6
n = 1 approx
So molecular formula is same as empirical formula
C₅H₁₀O is molecular formula .
Answer:
Your strategy here will be to use the molar mass of potassium bromide,
KBr
, as a conversion factor to help you find the mass of three moles of this compound.
So, a compound's molar mass essentially tells you the mass of one mole of said compound. Now, let's assume that you only have a periodic table to work with here.
Potassium bromide is an ionic compound that is made up of potassium cations,
K
+
, and bromide anions,
Br
−
. Essentially, one formula unit of potassium bromide contains a potassium atom and a bromine atom.
Use the periodic table to find the molar masses of these two elements. You will find
For K:
M
M
=
39.0963 g mol
−
1
For Br:
M
M
=
79.904 g mol
−
1
To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements
M
M KBr
=
39.0963 g mol
−
1
+
79.904 g mol
−
1
≈
119 g mol
−
So, if one mole of potassium bromide has a mas of
119 g
m it follows that three moles will have a mass of
3
moles KBr
⋅
molar mass of KBr
119 g
1
mole KBr
=
357 g
You should round this off to one sig fig, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs
mass of 3 moles of KBr
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
360 g
a
a
∣
∣
−−−−−−−−−
Explanation:
<em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em><em> </em><em>3</em><em>6</em><em>0</em><em> </em><em>g</em><em> </em>
Answer:
4 - 1 - 3 - 2 - 6 - 5
Explanation:
During an engineering process, first, we need to identify the problem, or the need because the process only will occur because of some need. Then, it's necessary to know as much as possible about the problem and the things that already exist or already were tested to solve it. Knowing the background will make the work easy.
After that, it's necessary to plan the things we'll do, knowing the costs, the time needed for activities, how many people will be necessary for each step, etc. It's really important to make a plan. Then, do the work, following the plan. Thus, the process must be tested. During the test of the results, some problems must be found, so it's time to evaluate and redesign the process, to solve these problems found.
Answer : The volume of the cube is, 
Solution : Given,
Density of nickel = 
Number of nickel atoms = 
Molar mass of nickel = 58.7 g/mole
First we have to calculate the moles of nickel.
As, 
atoms form 1 mole of nickel
So, atoms form 
moles of nickel
The moles of nickel = 3.321 moles
Now we have to calculate the mass of nickel.
The mass of nickel = 194.94 g
Now we have to calculate the volume of nickel.
Therefore, the volume of the cube is,
