Assume you do not know which peak in GC chromatogram corresponds to each alkene (and you are not sure whether the one with lower boiling point will appear first). Propose how you can identify which peak corresponds to which alkene in terms of "spiking products"
Answers
Spiking involves gradually increasing the concentration of one specific compound, and looking for a rise in peak height. Here, you can identify which peak corresponds to the compound you spiked. You can do this for each alkene, to determine its peak and retention time.
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70 g/mL
0.7 g/mL
700 g/mL
Answers
Answer:
The answer is 7 g/mL
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 70 g
volume = 10 mL
We have
We have the final answer as
7 g/mL
Hope this helps you
2. Ba2SO4
3. BaSO4
4. Ba(SO4)2
Answers
Answer:
3. BaSO4
Explanation:
the math will show you, and to answer your commented question "How does this evidence indicate that the reaction has occurred between ions ?" :
The reaction of the ions can be checked by a number of methods. In a chemical reaction, the ions are in the aqueous solution. This means that they are in solution, dissolved with charge. During the reaction (ionic) there is an attraction between the positive and negative ions. This makes a change in the state of the reactants in the products. In addition, the reaction can result in an immediate observable change such as the formation of a solid or precipitation.
0.50 moles solute
1.0 mole solute
2.0 moles solute
Answers
M = moles of solute / liter solution
2 molarity = moles of solute / 0.5 liters of solution
moles of solute = 1 moles of solute
Answer : The correct option is, 1.0 mole
Explanation : Given,
Molarity of solution = 2.0 mole/L
Volume of solution = 0.50 L
Molarity : It is defined as the number of moles of solute present in one liter of solution.
Formula used :
Now put all the given values in the above formula, we get the moles of the solute.
Therefore, the number of moles of solute dissolved in the solution will be, 1.0 mole
a
[H+] = [ ? ] x 10!?)
Coefficient (green)
Exponent (yellow)
Enter
Answers
Solution:
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Answers
Answer: you cannot find its new volume by using these known values and Boyle's law because the temperature does not remain constant.
Explanation:
Boyle's law states that the volume of a fixed amount of gas, at a constant temperature, varies inversely with the pressure.
So, it is a condition that the temperature does not change.
For the wheater ballon case, as it travels through the atmosphere, the temperature at different altitudes will be different.
So, you might use other equation of states, such as the combined law, which does deal with changes in the three variables: volume, pressure, and temperature.
The mathematical formulation of Boyle's law is:
pV = constanjt ⇒ p₁ V₁ = p₂ V₂, at constant T.
The mathematical formulation of the combined law of gases is:
pV/T = constant ⇒ p₁ V₁ / T₁ = p₂ V₂ / T₂, for a fixed amount of gas, then it might work for the weather ballon (if you know the initial and end temperatures).
You cannot find the new volume by using initial volume of the weather balloon and air pressure ai its initial and final altitudes and Boyle’s law because the given values are not the same. Boyle’s law holds for the pressure and volume of the GAS at constant temperature. Here you are given the air pressure outside the weather balloon not the inside of the balloon. They have different gases and so it would not apply.
Answers
1.) Determine the grams of carbon and hydrogen from the given data.
C: 23.98 g x (12.011 g / 44.01 g) = 6.54 g C
H: 4.91 g x (2.0158 g / 18.02 g) = 0.55 g H
Determine the grams of oxygen in the sample by subtracting the mass of the compound given from the total mass solved earlier.
O: 10.0 - (6.54 + 0.55) = 2.91 g O
2) Convert the values in step 1 to moles.
C: 6.54 g / 12.01 g / mol = 0.54 mol
H: 0.55 g / 1.01 g/mol = 0.54 mol
O: 2.91 g / 16.00 g/mol = 0.18 mol
3) Divide each by the lowest value calculated in step 2
C: 0.54 mol / 0.18 mol = 3
H: 0.54 mol / 0.18 mol = 3
O: 0.18 mol / 0.18 mol = 1
Thus, the empirical formula is C3H3O.