Answer:
8 mins and 20 seconds i think
Explanation:
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.
Learn more about titration here: brainly.com/question/4225093
Answer:
Explanation:
250 cm^3 of 0.2 moldm-3 H2SO4 can be prepared from 150cm^3 of 1.0 moldm^-3 by dilution.
150cm^3 of the 1.0 moldm^-3 stock solution is measured out using a measuring cylinder and transferred into a 250 cm^3 standard volumetric flask and made up to mark. The resulting solution is now 250cm^3 of 0.2 moldm-3 H2SO4.
The volume in liters occupied by 22.6 g of I₂ gas at STP is 1.99 L (answer A)
<u><em>calculation</em></u>
Step: find the moles of I₂
moles= mass÷ molar mass
from periodic table the molar mass of I₂ is 253.8 g/mol
moles = 22.6 g÷253.8 g/mol =0.089 moles
Step 2:find the volume of I₂ at STP
At STP 1 moles =22.4 L
0.089 moles= ? L
<em>by cross multiplication</em>
={ (0.089 moles x 22.4 L) /1 mole} = 1.99 L
To answer this question, you need to know <span>Graham's Law of Effusion/Diffusion formula. In this formula, the rate of diffusion/effusion would be influenced by the mass. As the molecule has bigger mass, the rate should be slower because it will be harder to pass the membrane. The calculation should be:</span>
<span>Rate 1 / Rate 2 = √[M2/M1]
</span>4.11/1= √[M2/2]
M2=33.78 g/mol
