Mass of SiC = 2 g
<h3>Further explanation</h3>
Given
Reaction
SiO₂(s) + 3C(s) → SiC(s) + 2CO(g)
3.00 g of SiO₂
4.50 g of C
Required
mass of SiC
Solution
mol SiO₂ (MW=60,08 g/mol) :
= 3 g : 60.08 g/mol
= 0.0499
mol C(Ar = 12 g/mol) :
= 4.5 g : 12 g/mol
= 0.375
mol : coefficient of reactants =
SiO₂ : 0.0499/1 = 0.0499
C : 0.375/3 = 0.125
SiO₂ as a limiting reactant(smaller ratio)
Mol SiC based on mol SiO₂ = 0.0499
Mass SiC :
= mol x MW
= 0.0499 x 40,11 g/mol
= 2 g
The pressure of the carbon dioxide will be 0.09079 atm.
<h3>What is partial pressure?</h3>
The pressure exerted by the individual gas is known as partial pressure.
The partial pressure is given as
In a mixture of carbon dioxide and oxygen, 40.0% of the gas pressure is exerted by oxygen.
If the total pressure is 115 mmHg.
The total pressure in atm will be
P = 115 mmHg
P = 0.15132 atm
We have
Then the pressure of the carbon dioxide will be 0.09079 atm.
More about the partial pressure link is given below.
brainly.com/question/13199169
#SPJ1
A refracting telescope uses convex lenses to gather a large amount of light and focus it onto a small area while a reflecting telescope the mirror instead of an objective lens they are also similar because like the lenses in a refracting telescope the mirror in a reflecting telescope focuses a large amount of light onto a small area.
To determine the volume of both concentration of vinegar, we need to set up two equations since we have two unknowns.
For the first equation, we do a mass balance:
mass of 100% vinegar + mass of 13% vinegar = mass of 42% vinegar
Assuming they have the same densities, then we can write this equation in terms of volume.
V(100%) + V(13%) = V(42%)
we let x = V(100%)
y = V(13%)
x + y = 150
For the second equation, we do a component balance:
1.00x + .13y = 150(.42)
x + .13y = 63
The two equations are
x + y = 150
x + .13y = 63
Solving for x and y,
x = 50
y = 100
Therefore, you need to mix 50 mL of the 100% vinegar and 100 mL of the 13% vinegar.
Answer: 4-allylanisole
Explanation: The doublets behind the 7 ppm belongs to the
para-substituted benzene ring. The three single-proton multi-plets around 5−6 ppm predicts that there has to be a single subsituted alkene group
A single plus a doublet around 3-4 ppm belongs to CH3 and CH2 Groups as they could be attached to the subsituted alkene group.
Moreover the interpretation of the NMR that there is no peak with a higher intensity for >180 ppm represents an absence of Carbonyl group.
The Predicted Number is attached from a chemical database along with their peaks information
