Answer: There are 78.26 mL of a 0.0023M strontium hydroxide solution are needed to completely react 15.0 mL of 0.012M hydrochloric acid.
Explanation:
Given:
= 0.0023 M,
= ?
= 15.0 mL,
= 0.012 M
Formula used to calculate volume of strontium hydroxide solution is as follows.

Substitute the values into above formula as follows.

Thus, we can conclude that there are 78.26 mL of a 0.0023M strontium hydroxide solution are needed to completely react 15.0 mL of 0.012M hydrochloric acid.
Answer:
18.4 g
Explanation:
M(He) = 4.0 g/mol
4.6 mol * 4.0 g/ 1 mol = 18.4 g
Answer: The value of equilibrium constant for new reaction is
Explanation:
The given chemical equation follows:
The equilibrium constant for the above equation is 
We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:
The equilibrium constant for this reaction will be the cube of the initial reaction.
If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.
The value of equilibrium constant for reverse reaction is:
Hence, the value of equilibrium constant for new reaction is
Answer:
A unit for measuring work is called the foot-pound
Consequently, the ion with the greatest nuclear charge (Al 3 +) is the smallest, and the ion with the smallest nuclear charge (N 3−) is the largest. The neon atom in this isoelectronic series is not listed in Table 2.8.3, because neon forms no covalent or ionic compounds and hence its radius is difficult to measure.