Answer:
3056.25g
Explanation:
Problem here is to find the mass of Cu(NO₃)₂ to weigh out to make for the number of moles.
Given;
Number of moles = 16.3moles
Unknown:
Mass of Cu(NO₃)₂ = ?
Solution:
To find the mass of Cu(NO₃)₂, use the expression below;
Mass of Cu(NO₃)₂ = number of moles x molar mass
Let's find the molar mass;
Cu(NO₃)₂ = 63.5 + 2[14 + 3(16)]
= 63.5 + 2(62)
= 187.5g/mol
Mass of Cu(NO₃)₂ = 16.3 x 187.5 = 3056.25g
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.
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Explanation:
Butanol (С4Н9OH)
Structural formula is :-
CH3-CH2-CH2-CH2-OH
Answer:
6.82 kg
Explanation:
Given that the amount of water is 15L and we know that the density of water is ≈ 1kg/L. The mass of water is given by mass = volume x density, i.e,
mass = 15 x 1 = 15 kg. Also the specific heat capacity of water is 4.186 KJ/kg.
The sublimation enthalpy of dry ice is 571 KJ/kg.
Now, the amount of heat lost by water is entirely used up for the sublimation (conversion from soild to gas) of dry ice. And the heat (Q) lost by water is given as : Q = mCΔT, where m is the mass of water, C the specific heat capacity of water and ΔT the change in temperature.
Here, Q = 15 x 4.186 x (90 - 28) = 3892.98 KJ.
This amount of heat is taken up by the dry ice for its sublimation. Also the energy taken by dry ice (Q') for its sublimation is given by: Q' = m'L', where m' is the mass of dry ice, L' is the latent heat of sublimation (i.e, the amount of heat required per kg of a substance to sublime) of dry ice amd L' = 571 KJ/kg.
Now, Q' =m'L' = heat lost by water = 3892.98KJ.
And, m'L' = m' x 571 KJ/kg = 3892.98 KJ. (Dividing with 571)
Therefore, m' = 6.82 kg.
and any derivation of it, the energy must
be in JOULES. We have the energy in eV.
So this will be a two step question. First,
convert the energy to joules (or use the
number from our equation sheet) then
find the wavelength. Once you are done
with that, pick the correct answer below.
8.29x10^-16 m
3.75x10^26 m
2.48x10^-7 m
3.98x10^-26 m
bro this is too hard man can do just quit
Answer:
c
Explanation:
would be produced from the
complete reaction of 83.7 g carbon
monoxide?
Fe2O3 + 3CO → 2Fe + 3CO2
131.6 grams of carbon dioxide would be produced from the complete reaction of 83.7 g carbon monoxide.
The balanced chemical equation is given below.
Fe2O3 + 3CO → 2Fe + 3CO2
Calculation,
Since, 28g of carbon dioxide produces 44g of carbon monoxide.
So, 83.7 g of carbon dioxide produces 44×83.7/28 grams
83.7 g of carbon dioxide produces 131.6 grams
The symbolic representation of chemical reaction in which reactant represents in left side and product represents in right side is called chemical equation.
To learn more about chemical equation here.
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Answer:131.6 g
Explanation: