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snow_lady [41]
3 years ago
15

An unknown compound contains 38.7 % calcium, 19.9 % phosphorus and 41.2 % oxygen. what is the empirical formula of this compound

?
Chemistry
1 answer:
Softa [21]3 years ago
8 0

Answer: Ca3(PO4)2

Explanation:

  • We assume that the sample is 100.0 g.
  • The sample contains 38.7 g of Ca, 19.9 g of P, and 41.2 g of O as the proportions show.
  • Then we can calculate the number of moles of each component (n = m/atomic mass).
  • The number of moles of Ca = 38.7/40.078 = 0.96 mol.
  • The number of moles of P = 19.9/30.97 = 0.64 mol.
  • The number of moles of O = 41.2/15.99 = 2.75 mol.
  • Now, we can get the molar ratios of different components (Ca, P, and O) in the sample by dividing the number of moles of each component by the lower number of moles, that we should divide the number of moles by (0.64).
  • Ca: P: O = (0.96/0.64) : (0.64/0.64) : (2.75/0.64) = 1.5 : 1 : 4.
  • To avoid the fraction of the ratios, we can multiply all ratios by 2.0.
  • Now, the ratio of Ca : P : O will be 3 : 2 : 8.
  • That main the empirical formula of the compound is Ca3P2O8 which can be expressed as calcium phosphate (Ca3(PO4)2).
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If the fugacity of a pure component at the conditions of an ideal solution is 40 bar and its mole fraction is 0.4, what is its f
yaroslaw [1]

Answer : The fugacity in the solution is, 16 bar.

Explanation : Given,

Fugacity of a pure component = 40 bar

Mole fraction of component  = 0.4

Lewis-Randall rule : It states that in an ideal solution, the fugacity of a component is directly proportional to the mole fraction of the component in the solution.

Now we have to calculate the fugacity in the solution.

Formula used :

f_i=X_i\times f_i^o

where,

f_i = fugacity in the solution

f_i^o = fugacity of a pure component

X_1 = mole fraction of component

Now put all the give values in the above formula, we get:

f_i=0.4\times 40\text{ bar}

f_i=16\text{ bar}

Therefore, the fugacity in the solution is, 16 bar.

4 0
3 years ago
What do u need to know to describe the velocity of an object?
Alex_Xolod [135]
To determine the velocity of an object, you need to know the displacement and the change in time. 

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5 0
3 years ago
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I want to know the steps.
Artyom0805 [142]

The answer for the following problem is described below.

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

Explanation:

Given:

enthalpy of combustion of glucose(ΔH_{f} of C_{6}H_{12} O_{6}) =-1275.0

enthalpy of combustion of oxygen(ΔH_{f} of O_{2}) = zero

enthalpy of combustion of carbon dioxide(ΔH_{f} of CO_{2}) = -393.5

enthalpy of combustion of water(ΔH_{f} of H_{2} O) = -285.8

To solve :

standard enthalpy of combustion

We know;

ΔH_{f}  = ∈ΔH_{f} (products) - ∈ΔH_{f} (reactants)

C_{6}H_{12} O_{6} (s) +6 O_{2}(g) → 6 CO_{2} (g)+ 6 H_{2} O(l)

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

ΔH_{f} = 6 (-393.5) + 6(-285.8)  - 0 + 1275

ΔH_{f} = -2361 - 1714 - 0 + 1275

ΔH_{f} =-2800 kJ

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

7 0
3 years ago
How many moles of FeCl3 could be produced from 6.1 moles of Cly?<br> 2 Fe + 3Cl2 —&gt; 2 FeCl2
Black_prince [1.1K]

Answer:

4.1 moles of FeCl₃

Explanation:

The reaction expression is given as shown below:

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Number of moles of Cl₂  = 6.1moles

So;

  We know that from the balanced reaction expression:

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Therefore    6.1moles of Cl₂ will produce \frac{6.1 x 2}{3}   = 4.1 moles of FeCl₃

The number of moles is 4.1 moles of FeCl₃

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Zigmanuir [339]

Answer:

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