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3 years ago

What is the concentration of an unknown Mg(OH)2 solution if it took an average of 15.4mL of

Chemistry

1 answer:

vova2212 [387]3 years ago

7 0

Answer:

0.077M

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2HCl + Mg(OH)2 —> MgCl2 + 2H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 2

The mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question.

Concentration of base Cb =...?

Volume of base (Vb) = 10mL

Concentration of acid (Ca) = 0.1M

Volume of acid (Va) = 15.4mL

Step 3:

Determination of the concentration of the base, Mg(OH)2.

The concentration of the base can be obtained as follow:

CaVa/CbVb = nA/nB

0.1 x 15.4 /Cb x 10 = 2/1

Cross multiply to express in linear form

Cb x 10 x 2 = 0.1 x 15.4

Divide both side by 10 x 2

Cb = (0.1 x 15.4) /(10 x 2)

Cb = 0.077M

Therefore, the concentration of the base, Mg(OH)2 is 0.077M

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By what factor would a scuba diver’s lungs expand if she ascended rapidly to the surface from a depth of 125 ft without inhaling

vova2212 [387]

It is found that scuba diver’s can safely ascend up to 52.53 ft without Breathing out.

By what factor would a scuba diver’s lungs expand if she ascended rapidly to the surface from a depth of 125 ft without inhaling or exhaling then

$p_{1}$ =patm+ $pH_{2} O$

$p_{1}$ = 101325+ρ $gh_{1}$

It is given that height is 125ft. Put the value of h in above formula:

h1 =125ft=38.1m

ρ=1.04g/mL=1040kg/ $m^{3}$

g=9.81

$p_{1}$ =101325Pa+388711.44

$p_{1}$ =490036.44‬Pa

$p_{2}$ =p atm =101325Pa

It is known that volume and pressure can be expressed as:

V*P=const.

where, V is volume and P is pressure.

Now,

$V_{1} *p_{1}$ = $V_{2} *p_{2}$

$V_{2} /V_{1}$ = $p_{2} /p_{1}$

$V_{2} /V_{1}$ =490036.44/101325

$V_{2} /V_{1}$ =4.84

Assume constant temperature

d of seawater = 1.04 g/mL; d of Hg = 13.5 g/mL.

now $p_{1}$ =p atm+ $pH_{2} O$ =490036.44Pa

V*p=const

where, V is volume and P is pressure.

Now,

$V_{1} *p_{1}$ = $V_{2} *p_{2}$

$V_{2} /V_{1}$ = $p_{2} /p_{1}$

$V_{2} /V_{1}$ =490036.44/X

$p_{2}$ =490036.44pa/(V2/V1) =326690.96Pa

$p_{2}$ =patm +p $H_{2} O$

$p_{2}$ =101325Pa+ρ $gh_{2}$

326690.96Pa=101325Pa+ρ $gh_{2}$

ρgh1 =151987.5-101325=225365.96‬‬Pa

ρ=1,04g/mL=1040kg/m3

g=9.81

$h_{2}$ =225365.96/‬ρ∗g

$h_{2}$ =225365.96 / ‬1040∗9.81

$h_{2}$ =22.09m= 72.47ft

ΔH= $H_{1} -H_{2}$

=125-72.47

=52.53ft

So she can safely ascend up to 52.53 ft without Breathing out

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