The average atomic mass if the element above is calculated by the sum of the product of the isotope abundance and its atomic mass unit. It is expressed as:
Average atomic mass = Σ xi(Mi)
<span>Average atomic mass = (.7547 x 248.7) + (.2453 x 249.4) = 248.87
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Answer: Ca(OH)2 (aq) + H2SO4 (aq) ----------> CaSO4(aq) + 2H2O(l)
Explanation:
Since this is a neutralization reaction, the end product would be salt and water. In this equation Calcium will displace hydrogen from the acid because it is more reactive, resulting in the formation of CaSO4 (salt), while the displaced H2 molecule combines with OH molecules to form water.
The equation of the reaction is thus;
Ca(OH)2 (aq) + H2SO4 (aq) ----------> CaSO4(aq) + H2O(l), in other to balance it, we add ''2'' to the water molecule in the right hand side of the equation.
Balance equation is
Ca(OH)2 (aq) + H2SO4 (aq) ----------> CaSO4(aq) + 2H2O(l)
56km/s is 201600m/hr to its equivalent.
Assuming the concentration of stock solution is 50% sodium phosphate buffer solution, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.
<h3>What volume of a stock Sodium phosphate buffer and water is needed to 12 mL of 25% sodium phosphate buffer of pH 4?</h3>
The process of preparing solutions from stock solutions of higher concentration is known as dilution.
Dilution is done with the aid of the dilution formula given below:
where
- C1 is the concentration of stock solution
- V1 is the volume of stock solution required to prepare a diluted solution
- C2 is the concentration of the diluted solution prepared
- V2 is the final volume of the diluted solution
From the data provided:
C1 is not given
V1 is unknown
C2 = 25%
V2 = 12 mL
- Assuming C1 is 50% solution
Volume of stock, V1, required is calculated as follows:
V1 = C2V2/C1
V1 = 25 × 12 /50
V1 = 6 mL
Therefore, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.
Learn more about dilution formula at: brainly.com/question/7208546
Answer:
H_2O + 2CrO_4^2- + 3SO_3^2- -> 3SO_3^2- + 2CrO_2^- + 2OH^-
Explanation:
Reduction half reaction
2H_2O + CrO_4^2- + 3e -> CrO_2^- + 4OH^-
Oxidation half reaction
2OH^- + SO_3^2- -> SO_4^2- + H_2O + 2e
Balanced overall equation
H_2O + 2CrO_4^2- + 3SO_3^2- -> 3SO_3^2- + 2CrO_2^- + 2OH^-
