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Reil [10]
3 years ago
10

A compound is found to be 38.76% calcium, 19.97% phosphorous and 41.28% oxygen. What is the empirical formula for this compound?

Chemistry
1 answer:
motikmotik3 years ago
3 0

The empirical formula is Ca₃P₂O₈.

<em>Assume</em> that you have 100 g of the compound.

Then you have 38.76 g Ca, 19.97 g P, and 41.28 g O.

Now, we must convert these <em>masses to moles</em> and <em>find their ratio</em>s.

If the number in the ratio are not close to integers, you <em>multiply them by a numbe</em>r that makes them close to integers.

From here on, I like to summarize the calculations in a table.

<u>Element</u>  <u>Mass/g</u>     <u> Moles  </u>      <u> </u><u>Ratio </u>    <u>   ×2    </u>   <u>Integers</u>  

     Ca       38.76      0.967 07     1.4998   2.9995         3

     P         19.97       0.644 82     1             2                   2

     O        41.28       2.580 0       4.0011   8.0023          8

The empirical formula is Ca₃P₂O₈.


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In a neutral solution the concentration of _____. a. hydrogen ions is less than the concentration of hydroxide ions b. water mol
Georgia [21]

Answer:

In a neutral solution the concentration of hydrogen ions is equal to the concentration of hydroxide ions

Explanation:

When:

[H⁺] > [OH⁻]  the pH is acid, so the solution is practically acid.

When

[H⁺] < [OH⁻], the pH is basic, so the solution is practically basic

In neutral solutions:

[H⁺] = [OH⁻], so the pH is neutral (7)

pH > 7 → BASIC SOLUTIONS

pH < 7 → ACID SOLUTION

5 0
4 years ago
How many grams of iron metal do you expect to be produced when 245 grams of an 80.5 percent by mass iron (II) nitrate solution r
NNADVOKAT [17]
2Al + 3Fe(NO₃)₂ = 3Fe + 2Al(NO₃)₃

m=245 g
w=0.805 (80.5%)
M{Fe(NO₃)₂}=179.857 g/mol
M(Fe)=55.847 g/mol

1. the mass of salt in solution is:
m{Fe(NO₃)₂}=mw

2. the proportion follows from the equation of reaction:
m(Fe)/3M(Fe)=m{Fe(NO₃)₂}/3M{Fe(NO₃)₂}

m(Fe)=M(Fe)m{Fe(NO₃)₂}/M{Fe(NO₃)₂}

m(Fe)=M(Fe)mw/M{Fe(NO₃)₂}

m(Fe)=55.847*245*0.805/179.857= 61.24 g


3 0
3 years ago
Read 2 more answers
The modern model of the atom shows that electrons are:
Lina20 [59]
Electrons are orbiting the nucleus in the fxed way paths located in solid sphere
4 0
3 years ago
A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in
In-s [12.5K]

Answer:

3.6667

Explanation:

<u>For helium gas:</u>

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 3.0 L

V₂ = 9.0 L

P₁ = 5.6 atm

P₂ = ?

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{5.6}\times {3.0}={P_2}\times {9.0} atm

{P_2}=\frac {{5.6}\times {3.0}}{9.0} atm

{P_1}=1.8667\ atm

<u>The pressure exerted by the helium gas in 9.0 L flask is 1.8667 atm</u>

<u>For Neon gas:</u>

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 4.5 L

V₂ = 9.0 L

P₁ = 3.6 atm

P₂ = ?

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{3.6}\times {4.5}={P_2}\times {9.0} atm

{P_2}=\frac {{3.6}\times {4.5}}{9.0} atm

{P_1}=1.8\ atm

<u>The pressure exerted by the neon gas in 9.0 L flask is 1.8 atm</u>

<u>Thus total pressure = 1.8667 + 1.8 atm = 3.6667 atm.</u>

6 0
3 years ago
The freezing-point depression of a 0.100 m MgSO4 solution is 0.225°C. Determine the experimental van't Hoff factor of MgSO4 at t
Andrews [41]

<u>Answer:</u> The experimental van't Hoff factor is 1.21

<u>Explanation:</u>

The expression for the depression in freezing point is given as:

\Delta T_f=iK_f\times m

where,

i = van't Hoff factor = ?

\Delta T_f = depression in freezing point  = 0.225°C

K_f = Cryoscopic constant  = 1.86°C/m

m = molality of the solution = 0.100 m

Putting values in above equation, we get:

0.225^oC=i\times 1.86^oC/m\times 0.100m\\\\i=\frac{0.225}{1.86\times 0.100}=1.21

Hence, the experimental van't Hoff factor is 1.21

7 0
3 years ago
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