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3 years ago

Calculate the percent composition of Ca3P2

Chemistry

2 answers:

Afina-wow [57]3 years ago

8 0

Calculate the molar mass of Ca3P2 in grams per mole or search for a chemical formula or substance.

torisob [31]3 years ago

7 0

Answer:

Ca - 66%, P - 34%

Explanation:

So, this is the formula we can use to find the amount of each element:

Element count * Atomic mass = Mass

Plug in our elements for this:

Ca - 3*40.078=120.234

P - 2*30.973=61.946

Now, to find the percentage of mass, we must find total mass, and divide the two elements mass count by this total mass:

120.234+61.946=182.18

Now divide each element mass by the total mass:

Ca - 120.234/182.18=0.6599(Round to 0.65)

P - 61.946/182.18=0.34002(Round to 0.34)

Then multiply both numbers by 100 to get the percentage:

Ca - 65%

P - 34%

So these our your two answer!

Hope this helps!

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Iron fluoride (FeF2) dissociates according to the following equation:

melomori [17]

Answer:

S = 0.788 g/L

Explanation:

The solubility product (Kps) is an equilibrium solubization constant, which can be calculated by the equation:

$Kps = \frac{[product]^x}{[reagent]^y}$

Where x and y are the stoichiometric coefficients of the product and the reagent, respectively. Because of the aggregation form, the concentration of solids is always equal to 1 for use in this equation.

Analyzing the equation, we see that for 1 mol of $Fe^{+2}$ is necessary 2 mols of $F^-$ , so if we call "x" the molar concentration of $Fe^2$ , for $F^-$ we will have 2x, so:

$Kps = [Fe^{+2}].[F^-]^2\\\\2.36x10^{-6} = x(2x)^2\\\\2.36x10^{-6} = 4x^3\\\\x^3 = 5.9x10^{-7}\\\\x = \sqrt[3]{5.9x10^{-7}} \\\\x = 8.4x10^{-3} mol/L$

So, to calculate the solubility (S) of FeF2, which is in g/L, we multiply this concentration by the molar mass of FeF2, which is:

Fe = 55.8 g/mol

F = 19 g/mol

FeF2 = Fe + 2xF = 55.8 + 2x19 = 93.8 g/mol

So,

[tex]S = 8.4x10^{-3}x93.8

S = 0.788 g/L

4 0

4 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

$p_{1}$ =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

To know more about Scuba diver here

brainly.com/question/15430942

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6 0

1 year ago

The sturdy wall outside of a skyscraper could be most easily compared to

Flauer [41]

I think the answer is A because the job of a cell membrane is to provide protection to the plant. A cell membrane has a sturdy wall.

3 0

3 years ago

The more the energy, the larger the (a,b,c,d)?

Elanso [62]

Answer: D:wavelenght

Explanation: Students will understand that shorter wavelengths have higher frequency and energy.

3 0

3 years ago

Read 2 more answers

What is the (OH-) in a solution with a pOH of 6.48

inn [45]

Through manipulation of equations, we are able to obtain the equation:

$-pOH= log [ OH^{-}]$

Then we can transform the equation into:

$[ OH^{-}]= 10^{-pOH}$

Then we are able to plug in the pOH and directly get [OH-]:

$[ OH^{-}] = 10^{-6.48}$

$[ OH^{-}]=3.31* 10^{-7} M$

3 0

3 years ago