Answer:
3.59x10^21 molecules
Explanation:
1mole of a substance contains 6.02x10^23 molecules.
Therefore, 1mole of C8H18 will also contain 6.02x10^23 molecules
1mole of C8H18 = (12x8) +(18x1) = 96 + 18 = 114g.
1mole (i.e 114g) oh C8H18 contains 6.02x10^23 molecules.
Therefore, 0.68g of C8H18 will contain = (0.68 x 6.02x10^23)/114 = 3.59x10^21 molecules
The labeled diagram is given in the image attached.
As it can be seen from the image that freezing is when energy is removed from the system at 0 ⁰ while melting is when energy is added at 0⁰.
Also when energy is added at 100⁰C, it causes boiling while when it is removed at 100⁰C, it causes condensation.
Melting point of water is 0⁰C while boiling point is 100⁰C
Answer:
pH = 12.15
Explanation:
To determine the pH of the HCl and KOH mixture, we need to know that the reaction is a neutralization type.
HCl + KOH → H₂O + KCl
We need to determine the moles of each compound
M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl
M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH
The base is in excess, so the HCl will completely react and we would produce the same mmoles of KCl
HCl + KOH → H₂O + KCl
3 m 4 m -
1 m 3 m
As the KCl is a neutral salt, it does not have any effect on the pH, so the pH will be affected, by the strong base.
1 mmol of KOH has 1 mmol of OH⁻, so the [OH⁻] will be 1 mmol / Tot volume
[OH⁻] 1 mmol / 70 mL = 0.014285 M
- log [OH⁻] = 1.85 → pH = 14 - pOH → 14 - 1.85 = 12.15
The mass of nitrogen collected is mathematically given as
M-N2=0.025gram
<h3>What is the mass of nitrogen collected?</h3>
Question Parameters:
A sample weighing 2.000g
the liberated NH3 is caught in 50ml pipeful of H2SO4 (1.000ml = 0.01860g Na2O).
T=26.3c=299.3K
Pressure=745mmHg=745torr
Pressure of N2=745-25.2=719.8torr
Generally, the equation for the ideal gas is mathematically given as
PV=nRT
Therefore
719.8/760=45.6/1000=n*0.0821*299.3
n=0.00176*14
In conclusion, the Mass of N2
M-N2=0.00176*14
M-N2=0.025gram
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CH₇ is the empirical formula of the car fuel.
Explanation:
To find the empirical formula we use the following algorithm.
First divide each mass the the molar weight of each element:
for carbon 2.87 / 12 = 0.239
for hydrogen 3.41 / 2 = 1.705
And now divide each quantity by the lowest number which is 0.239:
for carbon 0.239 / 0.239 = 1
for hydrogen 1.705 / 0.239 = 7.13 ≈ 7
The empirical formula of the car fuel is CH₇.
I have to tell you that in reality this formula is wrong because is not possible to exist. However the algorithm for finding the empirical formula is right, the problem may reside in the amounts of carbon and hydrogen given.
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empirical formula
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