Answer:
We need 75 mL of 0.1 M NaOH ( Option C)
Explanation:
Step 1: Data given
Molarity of NaOH solution = 0.100 M
volume of 0.150 M CH3COOH = 50.00 mL = 0.05 L
Step 2: The balanced equation
CH3COOH + NaOH → CH3COONa + H2O
Step 3: Calculate moles of CH3COOH
Moles CH3COOH = Molarity * volume
Moles CH3COOH = 0.150 M * 0.05 L
Moles CH3COOH = 0.0075 moles
Step 4: Calculate moles of NaOH
For 1 mol of CH3COOH we need 1 mol of NaOH
For 0.0075 mol CH3COOH we need 0.0075 mole of NaOH
Step 5: Calculate volume of NaOH
volume = moles / molarity
volume = 0.0075 moles / 0.100 M
Volume = 0.075 L = 75 mL
We need 75 mL of 0.1 M NaOH
The Ksp of FeCO₃ is 3.45x10-11 and the Ksp of CdCO₃ is 8.7x10-12.
Ksp is a mathematical expression used to predict the behavior of a saturated solution. It is an abbreviation of the term “solubility product constant”, and is derived from the thermodynamic equilibrium equation that exists between a solid and its dissolved ions in a solution. Ksp is used to predict the solubility of a given ionic compound, and is a measure of the maximum amount of solute that can be dissolved in a given solvent at a given temperature and pressure.
To initiate precipitation, the concentration of CO²⁻₃ must be greater than or equal to the solubility product of the least soluble salt, which is CdCO₃.
Therefore, the concentration of CO²⁻₃ must be greater than or equal to 8.7x10-12 M.
To learn more about Ksp
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Iodine value is a measure of the degree of unsaturation in fats and oils. It is essentially the number of grams of iodine consumed by 100 g of fat. If the iodine number is in the range of 0-70 then it is a fat, any value above 70 is considered an oil.
Formula:
Iodine number = (ml of 0.1 N Thiosulphate blank- ml of 0.1N thiosulphate test) * 12.7 *100/1000* wt of sample
vol of thiosulphate required to titrate test sample (given oil) = 1 ml
wt of sample = 0.2 g
Information on the volume of thiosulphate required to titrate the blank solution is essential for calculation.
Iodine number = (X-1.0) * 12.7 * 100/1000* 0.2 = (X-1.0)*6.35
Answer:
To determine the value of Kp for the given equilibrium, we need to use the partial pressures of the gases involved.
In the balanced equation: 2 HI (g) ⇌ H₂ (g) + I₂ (g), the stoichiometric coefficients are 2, 1, and 1 respectively.
At equilibrium, the expression for Kp is given by:
Kp = (P(H₂) * P(I₂)) / (P(HI)²)
Using the provided partial pressures:
P(HI) = 1.9 atm
P(H₂) = 7.9 atm
P(I₂) = 2.3 atm
Substituting these values into the expression for Kp:
Kp = (7.9 * 2.3) / (1.9²)
Kp ≈ 19.5 / 3.61
Calculating the result:
Kp ≈ 5.4
Therefore, the value of Kp for the given equilibrium is approximately 5.4.
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Answer:
=16.49 L
Explanation:
Using the equation
P1= 0.6atm V1= 30L, T1= 25+273= 298K, P2= 1atm, V2=? T2= 273
P1V1/T1= P2V2/T2
0.6×30/298= 1×V2/273
V2=16.49L
Answer:
5.4347826087
Explanation:
The formula for density is Mass/Volume so you would do 250/46 to get the answer of 5.4347826087 grams per milliliter