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Assoli18 [71]
3 years ago
12

when 10.00 g of phosphorus reacts with oxygen, it produces 17.77 g of a phosphorus oxide. This phosphorus oxide was found to hav

e a molecular mass of approximately 220 g/mole. Determine the molecular formula.
Chemistry
1 answer:
Alex_Xolod [135]3 years ago
8 0

Answer:

Molecular formula = P₄O₆

Explanation:

P(s)         +        O₂(g)------------------------------------⇒ PₓOₙ (g)

10g                     (17.77-10)g                                    17.77g

10g                        7.77g                                          17.77g  (gramme ratio)

The molecular mass of Phosphorus  (P) = 31g/mole

The molecular mass of Oxygen atom (O) = 16g/mole

Mole ratio is given by:

P              :             O

10/31                    7.77/16

0.3226       :         0.4856                Mole ratio---------------------------- (1)

Divide (1)  through by 0.3226

 1                 :        1.5-------------------------------------------- (2)

From  (2), the empirical formula for Phosphorus oxide :

Empirical formula = P₁O₁.₅

                               =  PO₁.₅

The molecular formula can be calculated from below:

Since the molecular formula is a multiple of the empirical formula we have

Molecular formula = (PO₁.₅)ₙ----------------------------------- (3)

Since we are given the molecular mass of the oxide formed, we have:

(PO₁.₅)ₙ = 220-----------------------------(4)

[31 + (16 x 1.5)] x n = 220

[31 + 24]n = 220

55n =220

n = 4

Substituting into (3), we have :

Molecular formula = (PO₁.₅)₄

                               = P₄O₆

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A gas has density 2.41 g/liter at 25°C and 770 mm Hg. Calculate it's molecular mass (R = 0.0821 L atm.mol-1K-1 ​
Evgesh-ka [11]

Answer:

Explanation:

Given : Density - 2.41 g/liter  

Temperature - 25° C  

Pressure : 770 mm Hg  

R = 0.0821 L atm mol-¹K-¹

 

Find : Molecular mass of gas

Solution : Ideal gas equation with respect to density will be : PM = dRT. In the formula, P is pressure, M is molecular mass, d is density, R is gas constant and T is temperature.

Keeping the values in equation-  

Pressure : 770 mm Hg = 1 atm  

Temperature : 273 + 25 = 298 K

 

M = dRT/P  

M = (2.41*0.0821*298)/1  

M = 58.96 gram/mol

Thus, the molecular mass of gas is 58.96 gram/mol.

8 0
3 years ago
NEED HELP QUICKLY!!! How many moles are in each of the following?
oksano4ka [1.4K]

Answer: a. 0.26mol

b. 0.000479mol

c. 1.12mol

Explanation: Please see attachment for explanation

6 0
3 years ago
A) Compute the repeat unit molecular weight of polystyrene. B) Compute the number-average molecular weight for a polystyrene for
Andrej [43]

Answer:

Explanation:

In Polystrene, the molecular formula for the repeat unit = C_8H_8;

and the atomic weights of Carbon C = 12.01 g/mol

For Hydrogen, it is 1.01 g/mol

Hence, the repeat unit molecular weight is:

m = 8 (12.01 g/mol)+8(1.01 g/mol)

m = 96.08 g/mol + 8.08 g/mol

m = 104.16 g/mol

The degree of polymerization = no-average molecular weight/repeat unit molecular weight.

Mathematically;

DP = \dfrac{\overline M_n}{m}

\overline M_n= DP \times m

\overline M_n= 25000 \times 104.16 \ g/mol

\overline M_n= 2604000  \ g/mol

7 0
2 years ago
3. What mass of copper could be deposited from a copper(II) Sulphate solution using a current of 5.0 A over 100 seconds? ( F =96
arsen [322]

Mass of copper : 0.165 g

<h3>Further explanation</h3>

Given

5.0 A over 100 seconds

Required

Mass of copper

Solution

Faraday's law:

<em>The mass of the substance formed at each electrode is proportional to the electric current flowing in the electrolysis</em>

<em />\tt W=\dfrac{e.i.t}{96500}<em />

e = Ar / valence = eqivalent weight

i = current

t = time

W = weight

CuSO₄ ----> Cu²⁺ + SO₄²⁻

Cu ----> Cu²⁺ + 2e

e = Ar/2

= 63,5/2 = 31,75

\tt W=\dfrac{31.75\times 5\times 100}{96500}=0.165~g

8 0
2 years ago
Isotopes that decay with a nuclear break-up and emit a significant amount of energy are said to be ________.
podryga [215]

Answer to this is Radioactive isotopes.

Isotopes are the species of the same element having different atomic masses that means the number of protons remains the same but number of neutrons do differ. For example _{1}^{2}\textrm{H} and _{1}^{3}\textrm{H} are the two isotopes of Hydrogen (_{1}^{1}\textrm{H} ).

Radioactive isotopes are the isotopes which release some kind of energy in the form of alpha particles, beta particles or gamma radiation. Examples of each of the decay processes are :

Alpha Decay: In this decay one alpha particle having atomic mass 4 and atomic number 2 or we can say a He molecule will come out. _{Z}^{A}\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_{2}^{4}\alpha

Beta Decay: In this decay a \beta particle is emitted increasing the atomic number of the reactant by 1 unit.

_{Z}^{A}\textrm{X}\rightarrow _{Z+1}^{A}\textrm{Y}+_{-1}^{0}\beta

Gamma Radiation: In this type of reaction only radiation is emitted out which does not change the original molecule.

_{Z}^{A}\textrm{X}\rightarrow _{Z}^{A}\textrm{X}+\gamma\text{ radiation}

3 0
2 years ago
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