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sesenic [268]
3 years ago
12

A compound consisting of only phosphorous and oxygen atoms is 43.64 % phosphorus by mass.The molecular mass/formula weight of th

e compound is 283.88 amu. How many phosphorous atoms ar in a molecule of this compound? Please show all calc involved.
Physics
1 answer:
DanielleElmas [232]3 years ago
6 0
First, we calculate the mass of Phosphorous present:
283.88 x 0.4364
= 123.88 amu
Atomic mass of P is 31 amu

moles of P = mass / Ar
= 123.88 / 31
= 4.0 moles

We know that one mole of substance has 6.02 x 10²³ particles
Atoms of P = 4 x 6.02 x 10²³ 
= 2.41 x 10²⁴ atoms
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Read 2 more answers
If The density of this stainless steel is7.85 g/cm3,specific heatis 0.5 J/g.K, melting pointis 1673K, heat of fusion s0.260J/kg.
kodGreya [7K]

Answer:

\Delta H=687.4 J

Explanation:

Hello!

In this case, for this melting process, we can identify two sub-processes in order to take the stainless steel from solid to liquid:

1. Heat up from 298.15 K to 1673 K.

2. Undergo the phase transition.

Both process have an associated enthalpy as shown below:

\Delta H_1=1g*0.5\frac{J}{g*K} (1673K-298.15K)=687.4J

\Delta H_2=0.001kg*\frac{0.260J}{kg} =0.00026J

Therefore, the required heat is:

\Delta H=\Delta H_1+\Delta H_2\\\\\Delta H=687.4J+0.00026J\\\\\Delta H=687.4J

Notice the problem is not providing neither the mass or volume, that is why we assumed the mass is 1 g; however, it can be changed to the mass you are given.

Best regards!

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