Solid water is completely clear
Answer:
2Zn + PbCl4 → 2ZnCl2 + Pb
N.B: ZnCl2 is zinc chloride
Pb is lead
Explanation:
Zinc has a chemical symbol Zn while lead(IV) chloride has the chemical formula, PbCl4. Combining these two substances in a reaction will give rise to the following:
Zn + PbCl4 → ZnCl2 + Pb
However, this equation is not balanced. To balance the equation, we make sure that the atoms of each element on both sides of the reaction are equal. This is done as follows;
2Zn + PbCl4 → 2ZnCl2 + Pb
N.B: ZnCl2 is zinc chloride
Pb is lead
The structure of sodium benzoate and benzoic acid is shown in the image.
Sodium benzoate is a sodium salt of benzoic acid.
Benzoic acid is an organic compound and weak acid means it does not dissociates in water that means it is sparingly soluble in water. Whereas the sodium benzoate is an ionic compound and dissociates completely in water to give sodium ions and benzoate ions that means it is highly soluble in water.
Thus, through solubility of benzoic acid and sodium benzote in water we can show difference between benzoic acid and sodium benzote. As, benzoic acid is sparingly soluble and sodium benzoate is highly soluble in water.
Formula for Molarity is given as,
Molarity = moles / Volume of Solution in L ----(1)
Also,
Moles = mass / M.mass
Putting value of mole in eq. 1,
Molarity = (mass / M.mass) / Vol
Solving for mass,
Mass = Molarity × M.mass × Vol
Putting Values,
Mass = 0.15 mol/L × 40 g/mol × 0.5 L
Mass = 3 g
Result:
Weight 3 g of NaOH, add it to a 500 ml volumetric flask, and add distilled water to the point of 500 ml. The resulting solution will be 0.15 M.
Answer:
a. 50KCal
b. 400KCal
c. Same as (a) above
Explanation:
Given
To raise the temperature of 1kg of liquid water at 1°C requires 1KCal
To raise the temperature of 1kg of ice or water vapour by 1°C requires 0.5KCal
To melt 1kg of ice at 0°C requires 80KCal
To evaporate 1kg of liquid water sitting at 100°C requires 540KCal
a. How much heat is required to raise the temperature of 5 kg of liquid water by 20 C?
To raise the temperature of 5 kg of ice by 20°C requires:
5 kg * (0.5 kcal / kgC) * 20C
= 50 KCal
b. How much heat is required to melt 5 kg of ice at 0 C?
To melt an ice of 5 kg of ice at 0 C requires:
5 kg * (80 kcal / kg)
= 400 KCal
c. Same as (a) above