Answers
Final answer:
Group/Family 18 on the periodic table is called the noble gases.
Explanation:
Group/Family 18 on the periodic table is called the noble gases. The noble gases are a group of chemical elements that have full valence electron shells, which makes them stable and nonreactive. This group includes elements like helium, neon, argon, krypton, xenon, and radon.
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How many liters of the antifreeze ethylene glycol [CH2(OH)CH2(OH)] would you add to a car radiator containing 6.50 L of water if the coldest winter temperature in your area is -10.°C? (The density of ethylene glycol is 1.11 g/mL. Assume the density of water at -10.°C is 1.00 g/mL.)
Hi ! Could someone help me out with this last part to my chemistry practice ? Thank you !
What is the mass, in grams, of 1.20×1021 molecules of aspirin, c9h8o4?
Answers
the amount of time it take to decrease from 1.00 m to 0.25 m
the answer is 0.75 h
B) Jet Fuel
C) steel
Answers
Option-C, STEEL is not a product of the fractional distillation of petroleum.
Explanation:
Petroleum is the mixture of Hydrocarbons *carbon and hydrogen containing compounds) present beneath the Earth's surface. Petroleum is formed from the remains of animals and plants beneath earth's surface in an anaerobic conditions.
Petroleum contains from small hydrocarbons (gases) to medium (liquids) and long chain hydrocarbons (Solids). These hydrocarbons are separated from each other by Fractional Distillation method (separation due to difference in boiling points)
Gasoline is a derivative of one of the fraction of petroleum used in internal combustion engines.
Jet Fuel is also derived from Kerosene and Naphtha fractions of petroleum.
While, Steel is inorganic Alloy (mixture of metals) composed of mainly Iron, Carbon and other elements.
Answers
Answer:
4.93g are extracted
Explanation:
Partition coefficient (P) is defined as the ratio of solute dissolved in the organic solvent and the solute dissolved in the aqueous phase.
That is:
P = 7.5 = Concentration in dichloromethane / Concentration in water.
Knowing this, in the first extraction with 25mL of dichloromethane you will extract:
7.5 = (X/25mL) / (5g - X) / 100mL
Where X is the amount of compound A that is extracted.
7.5 = 100X / (125 - 25X)
937.5 - 187.5X = 100X
937.5 = 287.5X
3.26g of A are extracted in the first extraction.
In water will remain 5g - 3.26g = 1.74g
In the second extraction you will extract:
7.5 = (X/25mL) / (1.74g - X) / 100mL
7.5 = 100X / (43.5 - 25X)
326.25 - 187.5X = 100X
326.25 = 287.5X
1.13g are extracted in the second extraction.
And remain: 1.74g - 1.13g = 0.61g
In the third extraction you will extract:
7.5 = (X/25mL) / (0.61g - X) / 100mL
7.5 = 100X / (15.25 - 25X)
114.375 - 187.5X = 100X
114.375 = 287.5X
0.40g are extracted in the third extraction.
And remain: 0.61g - 0.40g = 0.21g
In the second extraction you will extract:
7.5 = (X/25mL) / (0.21g - X) / 100mL
7.5 = 100X / (5.25 - 25X)
39.375 - 187.5X = 100X
39.375 = 287.5X
0.14g are extracted in the fourth extraction.
Thus, after the three extractions you will extract: 0.14g + 0.40g + 1.13g + 3.26g = 4.93g are extracted
Final answer:
The process involves using the partitioncoefficient to determine how much of Compound A will prefer the dichloromethane solvent over the water. Following a calculation process through four rounds of extraction, it is concluded that approximately 4.999g of Compound A will be extracted using four 25mL portions of dichloromethane.
Explanation:
The partition coefficient of a compound is a measure of how much it prefers one solvent over another. Given that the partition coefficient of Compound A is 7.5 in dichloromethane with respect to water, we can predict how much of this compound could be extracted using four separate 25 mL portions of dichloromethane.
Here's the step-by-step calculation process:
- We start with 5 grams of Compound A in 100 mL of water. Given the partition coefficient, in the initial phase, 5/(7.5+1)=0.625g remains in water and 7.5/8.5*5=4.375g goes into the dichloromethane.
- After one extraction with 25ml of dichloromethane, the amount left in the water will be 0.625g*1/(7.5+1)=0.069g.
- After the second extraction: 0.069g*1/(7.5+1) = 0.008g.
- After the third extraction: 0.008g*1/(7.5+1) = 0.0009g.
- After the fourth extraction: 0.0009g*1/(7.5+1) = 0.0001g.
In total, around 4.999g of compound A will be extracted using four 25mL portions of dichloromethane.
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Answers
Answer:
The answer is 85.71 mL
Explanation:
The new volume can be found by using the formula for Boyle's law which is
where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Since we are finding the new volume
We have
We have the final answer as
85.71 mL
Hope this helps you
b. 3.35
c. 2.41
d. 1.48
e. 7.00
Answers
Answer:
b. 3.35
Explanation:
To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.
pH = pKa + log ([salt]/[acid]) (Eq. 01)
Where
pKa = -log(Ka) (Eq. 02)
[salt] = Molar concentration of salt produced as a result of titration
[acid] = Molar concentration of acid left in the solution after titration
Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:
HNO2 + KOH ⇆ H2O + KNO2 (Eq. 03)
This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.
Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles
Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles
As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.
Therefore
Amount of salt produced i.e [salt] = 0.0025 moles (Eq. 04)
Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)
Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:
pH= -log(4.5x10 -4) + log (0.0025/0.0025)
Solving above we get
pH = 3.35
Final answer:
The pH value in the titration flask after 25.00 mL of the 0.10 M KOH solution is added to 50.00 mL of 0.10 M HNO2 solution is 3.35.
Explanation:
The subject of this question is titration, which is a method used in chemistry to measure the concentration of an unknown solution. Given 50.00 mL of 0.10 M HNO2 (nitrous acid, Ka = 4.5 × 10-4), titrated with 0.10 M KOH (potassium hydroxide), we need to calculate the pH after 25.00 mL of the KOH solution is added.
First, we need to find the moles of the HNO2 and the KOH. Moles equals Molarity times Volume. So, for HNO2, it is 0.10 M * 0.050 L which equals 0.005 moles. For KOH, it is 0.10 M * 0.025 L which equals 0.0025 moles.
Then, subtract the moles of OH- from the moles of HNO2 to determine the concentration of HNO2 left, which is 0.005 moles - 0.0025 moles = 0.0025 moles. Divide this by the total volume of the solution (50.00 mL + 25.00 mL = 75.00 mL or 0.075 L to determine the new concentration of HNO2, 0.0025 moles / 0.075 L = 0.033 M. Then use the given Ka value with the equation [H+] = sqrt(Ka * [HNO2]) to get [H+].
To find acids' pH, we use the formula pH = -log[H+]. Use the calculated [H+] to find the pH.
Upon performing these calculations, the resulting pH value should be approximately 3.35 after 25.00 mL of the KOH solution is added, so the answer is (b) 3.35.
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