Answer: The concentration of solution is 0.342 M
Explanation:
To calculate the molarity of solution, we use the equation:
We are given:
Mass of solute (Sodium chloride) = 15 g
Molar mass of sodium chloride = 58.5 g/mol
Volume of solution = 750 mL
Putting values in above equation, we get:
Hence, the concentration of solution is 0.342 M
The equation for the reaction is is 2H2O = 2H2 + O2
Ratio of H to O in water = 2:1
Molar mass of water = 2*1.008 + 15.999 = 18.015
18.015 g gives 2*1.008 hydrogen
20 gives 2* 1.008 * 20 / 18.015
= 2.24 g hydrogen to the nearest hundredth.
Mass of CO₂ = 1.3 x 10⁵ kg
A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.
Reaction
2C₈H₁₈ + 25O₂⇒ 16CO₂ + 18H₂O
16 gallon = 60566,6 ml
b. 3.35
c. 2.41
d. 1.48
e. 7.00
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
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.
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.
#SPJ3
Answer: The value of change in internal energy of the system is, 40 J.
Explanation : Given,
Heat absorb from the surroundings = 12 J
Work done on the system = 28 J
First law of thermodynamic : It is a law of conservation of energy in which the total mass and the energy of an isolated system remains constant.
As per first law of thermodynamic,
where,
= internal energy = ?
q = heat absorb from the surroundings
w = work done on the system
Now put all the given values in this formula, we get the change in internal energy of the system.
Therefore, the value of change in internal energy of the system is, 40J.
Answer:
8
Explanation:
Here we're dealing with the root mean square velocity of gases. We'll provide the formula in order to calculate the root mean square velocity of a gas:
Here:
is the ideal gas law constant;
is the absolute temperature in K;
is the molar mass of a compound in kg/mol.
We know that the gas from the red container is 4 times faster, as it takes 4 times as long for the yellow container to leak out, this means:
We also know that the temperature of the red container is twice as large:
Write the ratio of the velocities and substitute the variables:
Then:
From here:
Then:
Considering Graham's Law of Effusion, and given that the temperature in the red container is twice that in yellow, the molar mass of the gas in the yellow container is 16 times that of the gas in the red container.
The question is about comparing the molar masses of the gases based on the rate at which they escape or effuse from two different containers. The key to this problem lies in understanding Graham's Law of Effusion, which states that the rate at which a gas effuses is inversely proportional to the square root of its molar mass.
Firstly, note that it is given that the red container takes 1/4th the time as yellow to effuse completely, meaning the gas in the red container effuses 4 times faster than the gas in the yellow container. Hence, the ratio of rates of effusion is 4:1
It is also given that the temperature in the red container is twice that in the yellow. Given the gases are in the same volume and pressure, by Graham's law, the ratio of molar masses (Myellow / Mred) would be the square of the ratio of their effusion rates, however when different temperatures are considered, it's the square of [ratio of their effusion rates x (Tred / Tyellow)].
So the ratio of the molar mass of the yellow container to the red would be (4*22)2 = 16, implying that the molar mass of the gas in the yellow container is 16 times that of the gas in the red container.
#SPJ3
OB. only the important elements that exist in the world.
OC. only the important compounds that exist in the world.
OA. all the known elements that exist in the world today.