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Dafna1 [17]
4 years ago
7

4. Given the chemical formulas of the following compounds, name each compound and state the rules applied to determine each name

. Score • RbF • CuO • (NH4)2C2O4 (Note: C2O4– is called oxalate.)
Chemistry
2 answers:
ollegr [7]4 years ago
5 0
1) RbF- rubidium fluoride
Rubidium fluoride is salt with ionic bond.
Rubidium is a cation which should be named first and followed by fluorine which is an anion.

2) CuO- copper(II) oxide
CuO is oxide which first goes name of metal and its oxidation state (+2 - II) at the end oxide.

3) (NH₄)₂C₂O₄- ammonium oxalate.
This is a salt compound and ammonium (NH₄⁺) is cation and the chemical formula for oxalate is C₂O₄^-2
Charra [1.4K]4 years ago
4 0
1) RbF is rubidium fluoride. Rubidium fluoride is salt with ionic bond. Rubidium is cation (goes first in name) and fluorine is anion (fluoride anion).
2) CuO is copper(II) oxide. CuO is oxide (compound<span> that contains at least one </span>oxygen<span> atom and one other element), first goes name of metal, his oxidation state (+2 - II) at the end oxide.
3) (NH</span>₄)₂C₂O₄ is ammonium oxalate. This compound is salt also, ammonium (NH₄⁺) is cation.
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OleMash [197]

Answer:

ΔH°f P4O10(s) = - 3115.795 KJ/mol

Explanation:

  • P4O10(s) + 6H2O(l) ↔ 4H3PO4(aq)
  • ΔH°rxn = ∑νiΔH°fi

∴ ΔH°rxn = - 327.2 KJ

∴ ΔH°f H2O(l) = - 285.84 KJ/mol

∴ ΔH°F H3PO4(aq) = - 1289.5088 KJ/mol

⇒ ΔH°rxn = (4)(- 1289.5088) - (6)(- 285.84) - ΔH°f P4O10(s) = - 327.2 KJ

⇒ ΔH°f P4O10(s) = - 5158.035 + 1715.04 + 327.2

⇒ ΔH°f P4O10(s) = - 3115.795 KJ/mol

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2 years ago
Three 6−l flasks, fixed with pressure gauges and small valves, each contain 6 g of gas at 276 k. flask a contains ch4, flask b c
Margaret [11]

First, please check the missing part in your question in the attachment.

a) So first, the Rank of pressure:

according to this formula PV = nRT and when n = m/Mw

PV = m/Mw * R*T

when we have the same mass m and the same V volume so P will proportional with the mole weight M as when the M is smaller the pressure will be greater 

when Mw of H2(A) = 2 g / Mw of He (B) = 4 g and Mw of CH4(C) = 16 g

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 (A) > (B) > (c)

B) The rank of average molecular kinetic energy:

when K = 3/2 KB T

when K is the average kinetic energy per molecule of gas 

and KB is Boltzmann's constant

and T is the temperature (K)

So from this equation, we can know that K only depends on T value, and when we have the T constant here for A, B, and C So the rank of K will be like the following:

∴ A = B = C

C) the rank of diffusion rate after the valve is opened:

according to this formula:

R2/R1 = √M1/M2

from this equation, we can see that diffusion is proportional to the reciprocal of the molecular mass M so,

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∴ the rank of diffusion:

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D) The rank of the Total kinetic energy of the molecules:

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A > B > C

e) the rank of density:

when ρ = m/ v 

and m is the mass & v is the volume and we have both is the same for A, B, and C

so the density also will be the same, ∴ the rank of the density is:

A = B = C

F) the rank of the collision frequency:

as the no.of molecules increase the collision frequency increase and depend also on the velocity and it's here the same.

∴ Collision frequency will only depend on the no.of molecules

we have no.of molecules of A > B > C as Mw A < B < C 

∴the rank of the collision frequency is:

A > B > C 

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