The structure that corresponds to methylpropylether is the structure CH3 - O - C3H7 in option C.
<h3>What is structure?</h3>
The structure of a compound is defined as the way in which the atoms in the compound are arranged. We know that a compound is composed of atoms and these atoms are arranged in particular patterns in space. This is the stereochemistry of the molecule.
Now we know that methylpropylether is a compound that contains the methyl group and the propyl group separated by a oxygen atom as is typical of all ethers.
As such, the structure that corresponds to methylpropylether is the structure CH3 - O - C3H7 in option C.
Learn more about methylpropylether:brainly.com/question/28047849
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I think B qualitative is the answer
The question in English is "<span>Determine the mass, in kg, of a material that is contained in a volume of 18L. It is known that the material density is 0.9 g/cm 3"
Answer:
</span>
We can use a simple
equation to solve this problem. <span>
d =
m/v</span><span>
<span>Where </span>d <span>is
the density, </span>m <span>is
the mass and </span>v is the volume.
d = </span>0.9<span> g/cm³
m = ?
v = </span>18 L = 18 x 10³ cm³<span>
By applying the equation,
<span> 0.9 g/cm³ = m / </span></span>18 x 10³ cm³<span>
m = 0.9 g/cm³ x </span>18 x 10³ cm³<span>
<span>
</span>m = 16200 g
m = 16.2 kg
Hence, the mass of
18 L of material is 16.2 kg.</span>
Answer:1.
1.Balanced equation
C4H10 + 9 02 ==> 5H20 +4CO2
2. Volume of CO2 =596L
Explanation:
1.Combustion of alkane is the reaction of alkanes with Oxygen. And the general equation for the combustion is;
CxHy +( x+y/4) O2 ==> y/2 02 + xCO2
Where x and y are number of carbon and hydrogen atoms respectively.
For butane (C4H10)
x=4 and y=10
Therefore
C4H10 + 9 02 ==> 5H20 +4CO2
2. Mass of butane = 0.360kg
Molar mass of C4H10 = ( 12×4 + 1×10)
= 48 +10=58g/mol= 0.058kg/mol
Mole = mass/molar mass
Mole = 0.360/0.058= 6.2moles
From the stoichiometric equation
1mole of C4H10 will gives 4moles of CO2
Therefore
6.2moles of C4H10 will gives 4 moles of 24.8 moles of CO2
Using the ideal gas equation
PV=nRT
P= 1.0atm
V=?
n= 24.8mol.
R=0.08206atmL/molK
T=20+273=293
V= 24.8 × 0.08206 × 293
V= 596L
Therefore the volume of CO2 produced is 596L
