When lewis structures are drawn, do you only show valence electrons??

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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6 0

2 years ago

Question 7 of 15

Crazy boy [7]

Answer: 0.4 moles

Explanation:

Given that:

Volume of gas V = 11L

(since 1 liter = 1dm3

11L = 11dm3)

Temperature T = 25°C

Convert Celsius to Kelvin

(25°C + 273 = 298K)

Pressure P = 0.868 atm

Number of moles N = ?

Note that Molar gas constant R is a constant with a value of 0.00821 atm dm3 K-1 mol-1

Then, apply ideal gas equation

pV = nRT

0.868atm x 11dm3 = n x (0.00821 atm dm3 K-1 mol-1 x 298K)

9.548 atm dm3 = n x 24.47atm dm3mol-1

n = (9.548 atm dm3 / 24.47atm dm3 mol-1)

n = 0.4 moles

Thus, there are 0.4 moles of the gas.

3 0

3 years ago

A chemist dissolves 867. mg of pure barium hydroxide in enough water to make up 170. mL of solution. Calculate the pH of the sol

Ilya [14]

Answer: The pH of the solution is 11.2

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

moles of $Ba(OH)_2$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{0.867g}{171g/mol}=0.00507mol$ (1g=1000mg)

Now put all the given values in the formula of molality, we get

$Molarity=\frac{0.00507\times 1000}{170}$

$Molarity=0.0298$

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pOH=-\log [OH^-]$

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^{-}$

According to stoichiometry,

1 mole of $Ba(OH)_2$ gives 2 mole of $OH^-$

Thus 0.0298 moles of $Ba(OH)_2$ gives = $\frac{2}{1}\times 0.0298=0.0596$ moles of $OH^-$

Putting in the values:

$pOH=-\log[0.0596]=2.82$

$pH+pOH=14$

$pH=14-2.82$

$pH=11.2$

Thus the pH of the solution is 11.2

8 0

3 years ago

Predict the products that will be formed by the reaction below. Al(s)+Na2SO4(s)—>

Ostrovityanka [42]

Answer:

Al₂(SO₄)₃ and Na

Explanation:

Al has a charge of +3, Na has a charge of +1 and SO₄ has a charge of -2. Since cations and anions will bond we know that Al will bond with SO₄ leaving Na by itself (since this is a single replacement reaction). When Al bonds with SO₄ it makes aluminum sulfate which is Al₂(SO₄)₃ and Na will be left by itself.

3 0

3 years ago

Predict what would be formed

Anarel [89]

Answer:

No

Explanation:

T6tgbv. 55678 r4fyx a

8 0

3 years ago