0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.
Potassium hydrogen phthalate, KHP, is a monoprotic acid often used to standardize NaOH solutions.
The balanced neutralization equation is:
NaOH(aq) + KHC₈H₄O₄(aq) ⇒ KNaC₈H₄O₄(aq) + H₂O(l)
- Step 1: Calculate the reacting moles of KHP.
0.212 g of KHP react. The molar mass of KHP is 204.22 g/mol.
0.212 g × 1 mol/204.22 g = 1.04 × 10⁻³ mol
- Step 2: Determine the reacting moles of NaOH.
The molar ratio of NaOH to KHP is 1:1.
1.04 × 10⁻³ mol KHP × 1 mol NaOH/1 mol KHP = 1.04 × 10⁻³ mol NaOH
- Step 3: Calculate the molarity of NaOH.
1.04 × 10⁻³ moles of NaOH are in 35.00 mL of solution.
[NaOH] = 1.04 × 10⁻³ mol / 35.00 × 10⁻³ L = 0.0297 M
0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.
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Baloon with 3 moles og oxygen at 1 atm.The temperature of the balloon is <u>4 Kelvin</u>.
An ideal gas is a theoretical gas composed of many randomly transferring factor particles that aren't difficult to interparticle interactions. the best gasoline idea is beneficial because it obeys the precise gas law, a simplified equation of country, and is amenable to evaluation under statistical mechanics.
An ideal gas is described as one for which both the extent of molecules and forces between the molecules are so small that they have got no effect at the behavior of the gas. The real gas that acts almost like a really perfect gasoline is helium. that is due to the fact helium, in contrast to maximum gases, exists as an unmarried atom, which makes the van der Waals dispersion forces as low as viable
Using the ideal gas equation:-
Given;
P₁ = 1 atm
V₁ = 100 L
n = 3
r = 8.314
T = PV/nR
= 1 × 100 / 3 × 8.314
= 4 K
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Answer:
I could create a slower reaction because the particles might be moving slower due to the cold. if it was warm there will be a faster reaction. similar to the elements movements in solids and liquids.
Answer:
0.9715 Fraction of Pu-239 will be remain after 1000 years.
Explanation:
Where:
= decay constant
=concentration left after time t
= Half life of the sample
Half life of Pu-239 = 
[
![\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B0.693%7D%7B24%2C000%20y%7D%3D2.8875%5Ctimes%2010%5E%7B-5%7D%20y%5E%7B-1%5D)
Let us say amount present of Pu-239 today = 
A = ?
![A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}](https://tex.z-dn.net/?f=A%3Dx%5Ctimes%20e%5E%7B-2.8875%5Ctimes%2010%5E%7B-5%7D%20y%5E%7B-1%5D%5Ctimes%201000%20y%7D)
0.9715 Fraction of Pu-239 will be remain after 1000 years.
Answer is: the maximum concentration of Pb²⁺ is 6.8·10⁻³ M.
Chemical reaction 1: PbCl₂(s) → Pb²⁺(aq) + 2Cl⁻(aq).
Chemical reaction 2: NaCl(aq) → Na⁺(aq) + Cl⁻(aq).
Ksp(PbCl₂) = 1.7·10⁻⁵.
c(NaCl) = c(Cl⁻) = 0.0500 M.
Ksp(PbCl₂) = c(Pb²⁺) · c(Cl⁻)².
c(Pb²⁺) = Ksp(PbCl₂) ÷ c(Cl⁻)².
c(Pb²⁺) = 1.7·10⁻⁵ M³ ÷ (0.0500 M)².
c(Pb²⁺) = 0.000017 M³ ÷ 0.0025 M².
c(Pb²⁺) = 0.0068 M = 6.8·10⁻³ M.
