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3 years ago

What is the mass of oxygen in 300 grams of carbonic acid (H2CO3)

Chemistry

1 answer:

DENIUS [597]3 years ago

7 0

mass of carbonic acid = 300g

molar mass of H2CO3 = 2H + C + 3 O

= 2 x 1.008+ 12.01 + 3 x 16

= 62.03g/mol

moles of H2CO3 = mass/Molar mass

= 300/62.03

= 4.8364 moles

1 mole H2CO3 has 3 moles Oxygen

4.8364 moles H2CO3 contains

= 3 x 4.8364 moles Oxygen = 14.509 moles Oxygen

moles = mass/Molar mass

mass of oxygen = moles x Molar mass of Oxygen

= 14.509 x 16

= 232.15g Oxygen

mass of oxygen in 300g of carbonic acid(H2CO3) = 232.15g

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calculation

volume of granite = (volume of cylinder after placing granite - volume of cylinder before placing granite

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Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy for the following redox

stealth61 [152]

Answer:

-973 KJ

Explanation:

The balanced reaction equation is;

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From;

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3 years ago

How many moles of Ca atoms are in 1 mol of CaSO4?

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Answer:

6.02 x 10²³ atoms

Explanation:

Given parameters:

Number of moles CaSO₄ = 1 mole

Unknown:

Number of Ca atoms in the given compound = ?

Solution:

The given compound is:

CaSO₄

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1 mole of any substance contains 6.02 x 10²³ atoms

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3 years ago

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

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$p_{1}$ =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

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Assume constant temperature

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

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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

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g=9.81

$h_{2}$ =225365.96/‬ρ∗g

$h_{2}$ =225365.96 / ‬1040∗9.81

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

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