Answer:
The answer to your question is letter A.
Explanation:
Isomers are molecules that have the same molecular formula but have a different structure. The molecule from which are looking an isomer has 5 carbons and 1 double bond. Then we need to look for another molecule with these components.
A.- This molecule has 5 carbons and 1 double bond, This structure is an isomer of the first one.
B.- This molecule has 3 carbons and 1 double bond, it's not an isomer of the first structure.
C. This molecule has 4 carbons and 1 triple bonds, it's not an isomer of the first structure.
D. This molecule has 5 carbons but it doesn't have any double bond, then it's not an isomer of the first structure.
Answer:
B.
Explanation:
Everything else is not a drawback, it is a benefit.
Answer:
0.0119
Explanation:
There was a part missing. I think this is the whole question:
<em>Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the concentration of A at equilibrium is 0.0153 M?</em>
A (aq) ⇌ 2B (aq) + C(aq)
<em>Remember to use correct significant figures in your answer. Do not include units in your response.</em>
First, we have to make an ICE Chart, which stands for initial, change and equilibrium. We will call "x" unknown concentrations.
A (aq) ⇌ 2B (aq) + C (aq)
I 0.0510 0 0
C -x +2x +x
E 0.0510-x 2x x
Since the concentration at equilibrium of A is 0.0153 M, we get
We can use the value of x to calculate the concentrations at equilibrium.
![[A]e = 0.0153 M \\[B]e = 2x = 2(0.0357) = 0.0714 M \\[C]e = x = 0.0357 M \\](https://tex.z-dn.net/?f=%5BA%5De%20%3D%200.0153%20M%20%5C%5C%5BB%5De%20%3D%202x%20%3D%202%280.0357%29%20%3D%200.0714%20M%20%5C%5C%5BC%5De%20%3D%20x%20%3D%200.0357%20M%20%5C%5C)
The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.
![Kc = \frac{[B]^{2} \times [C]}{[A]} = \frac{0.0714^{2} \times 0.0357}{0.0153} = 0.0119](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BB%5D%5E%7B2%7D%20%20%5Ctimes%20%5BC%5D%7D%7B%5BA%5D%7D%20%3D%20%5Cfrac%7B0.0714%5E%7B2%7D%20%20%5Ctimes%200.0357%7D%7B0.0153%7D%20%3D%200.0119)
The equilibrium constant for this reaction at equilibrium is 0.0119.
You can learn more about equilibrium here: brainly.com/question/4289021
The number of grams of Mgo that are produced is calculated as follows
calculate the moles of Mg used
that is moles =mass/molar mass
= 3.45 g/ 24.305g/mol = 0.142 moles
write the equation for reaction
that is 2 Mg + O2 = 2 MgO
by use of mole ratio between Mg to Mgo which is 2:2 this implies that the moles of Mgo is also 0.142 moles
mass of Mgo= moles x molar mass of MgO
molar mass of Mgo = 24.305 + ( 3x15.999) = 72.302 g/mol
mass= 0.142 mol x 72.302 g/mol= 10.27 grams
Answer:
SO3
Explanation:
Data obtained from the question include:
S = 40%
O = 59%
To obtain the empirical formula, do the following:
Divide the above by their molar mass as shown below:
S = 40/32 = 1.25
O = 59/16 = 3.69
Next, divide by the smallest as shown below:
S = 1.25/125 = 1
O = 3.69/1.25 = 3
Therefore, the empirical formula is SO3
