Answer:
2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.
Explanation:
Generally, moles of solute in solution before dilution must equal moles of solute after dilution.
By definition Molarity = moles solute/volume of solution in Liters
=> moles solute = Molarity x Volume (L)
Apply moles before dilution = moles after dilution ...
=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution
=> (M)(2.5L)before = (1.2M)(10.0L)after
=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate
The balanced equation for the above reaction is as follows;
2HCl + K₂SO₃ ---> 2KCl + H₂O + SO₂
stoichiometry of HCl to SO₂ is 2:1
number of moles of HCl reacted - 15.0 g / 36.5 g/mol = 0.411 mol
according to molar ratio
number of SO₂ moles formed - 0.411 mol /2 = 0.206 mol
since we know the number of moles we can find volume using ideal gas law equation
PV = nRT
where
P - pressure - 1.35 atm x 101 325 Pa/atm = 136 789 Pa
V - volume
n - number of moles - 0.206 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 325 K
substituting values in the equation
136 789 Pa x V = 0.206 mol x 8.314 Jmol⁻¹K⁻¹ x 325 K
V = 4.07 L
volume of SO₂ formed is 4.07 L
Answer:
When two single single bonds separated by a double bond (e.g C=C-C=C or C=C-C=O in the case of 2-cyclohexenone), the effect of resonance among those there bonds will be observed.
Explanation:
Since the Oxygen atom has higher electronegativity, it will cause the electrons in the resonance bonds 'flow' toward the Oxygen atom, so that the C=C will 'lose' some electron. The signal read for that bond will be different from other alkene structure.
Attachment is the resonance structure of 2-cyclohexene.
Answer:
The new force will be \frac{1}{100} of the original force.
Explanation:
In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.
That said, let's say that our equation for the initial force is:
![F = G\frac{m_1m_2}{R^2}The problem states that the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R](https://tex.z-dn.net/?f=F%20%3D%20G%5Cfrac%7Bm_1m_2%7D%7BR%5E2%7D%3C%2Fp%3E%3Cp%3EThe%20problem%20states%20%20that%20%20the%20distance%20decrease%20to%201%2F10%20of%20the%20original%20distance%2C%20this%20means%3A%3C%2Fp%3E%3Cp%3E%5Btex%5DR_2%20%3D%20%5Cfrac%7B1%7D%7B10%7DR)
And the force at this distance would be written in terms of the same equation:

Find the ratio between the final and the initial force:

Substitute the value for the final distance in terms of the initial distance:

Simplify:

This means the new force will be \frac{1}{100} of the original force.
Would all have the same ratio of elements.