Its an inelastic collision because the force from the bat causes it to bounce back. It is also an elastic force because catching the ball call for the energy of the ball to be deformed and restored into the mitt.

8 0

3 years ago

Calculate the pH of: (a) 0.1M HCl; (b) 0.1M NaOH; (c) 3 X 10% M HNO3; (d) 5 X 10-10 M HCIO.; and (e) 2 x 10-8 M KOH.

Shtirlitz [24]

Answer:

(a) $pH = -Log (0.1M) = 1$

(b) $pH = -Log (10^{-13}M) = 13$

(d) $pH = -Log (4.93x10^{-10}M) = 9.3$

(e) $pH = -Log (5^{-7}M) = 6.3$

Explanation:

To calculate de pH of an acid solution the formula is:

$pH = -Log ([H^{+}]) = 1$

were [H^{+}] is the concentration of protons of the solution. Therefore it is necessary to know the concentration of the protons for every solution in order to solve the problem.

(a) and (c) are strong acids so they dissociate completely in aqueous solution. Thus, the concentration of the acid is the same as the protons.

(b) and (e) are strong bases so they dissociate completely in aqueous solution too. Thus, the concentration of the base is the same as the oxydriles. But in this case it is necessary to consider the water autoionization to calculate the protons concentration:

$K_{w} =[H^{+} ][OH^{-}]=10^{-14}$

clearing the $[H^{+} ]$

$[H^{+} ]=\frac{10^{-14}}{[OH^{-}]}$

(d) is a weak base so it is necessary to solve the equilibrium first, knowing $Ka=3.24x10^{-8}$

The reaction is $HClO$ → $H^{+} + CO^{-}$ so the equilibrium is

$Ka=3.24x10^{-8}=\frac{x^{2}}{5x10^{-8}-x}$

clearing the <em>x</em>

${x^{2}={1.62x10^{-17}-3.24x10^{-8}x}$

$x=[H^{+}]=4.93x10^{-10}$

6 0

3 years ago

D( )-Camphor is treated with 5-methoxytryptamine in the presence of sodium cyanoborohydride to generate a new molecule, as a sin

Svetach [21]

The reducing agent can approach the carbonyl face of camphor by forming a one carbon bridge (known as an exo attack) or a two carbon bridge (termed endo).

The two resultant stereoisomers are known as isoborneol and borneol (from exo attack) (from endo attack). Gas chromatography (GC) analysis may be used to calculate the ratio of each isomeric alcohol in the mixture. Unfortunately, IR analysis does not permit this.

The stereochemistry of the reaction is regulated in stiff cyclic compounds like camphor and norcamphor by protecting one side of the carbonyl group from the reagent's assault. The hydrogen atom is added to the endo side, creating the exo alcohol isoborneol, while the methyl groups on the one-carbon bridge of camphor screen the approach of the hydride from the "top" or exo side of the two-carbon bridge. You will be asked to guess the main isomeric alcohol created by the norcamphor hydride reduction later in the lab report.

To view more about rational reaction, refer to:

brainly.com/question/20308523

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