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Angelina_Jolie [31]
4 years ago
14

Why does sulfur have a higher melting point than phosphorus

Chemistry
1 answer:
tiny-mole [99]4 years ago
5 0
Because phosphorus is more flammable than sulfur!

Have a nice day! :)

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Need help with this.
ankoles [38]
The answer would be D!
3 0
3 years ago
Read 2 more answers
How is the molecule of substance formed​
rosijanka [135]

Answer:

When atoms approach one another closely, the electron clouds interact with each other and with the nuclei. If this interaction is such that the total energy of the system is lowered, then the atoms bond together to form a molecule.

Explanation:

8 0
3 years ago
Find the concentration of H+ ions at a pH = 11 and pH = 6. Then divide the concentration of H+ ions at a pH = 11 by the of H+ io
andrew11 [14]

Explanation:

When pH of the solution is 11.

pH=-\log[H^+]

11=-\log[H^+]

[H^+]=1\times 10^{-11} M..(1)

At pH = 11, the concentration of H^+ ions is 1\times 10^{-11} M.

When the pH of the solution is 6.

pH=-\log[H^+]'

6=-\log[H^+]'

[H^+]'=1\times 10^{-6} M..(2)

At pH = 6, the concentration of H^+ ions is 1\times 10^{-6} M.

On dividing (1) by (2).

\frac{[H^+]}{[H^+]'}=\frac{1\times 10^{-11} M}{1\times 10^{-6}}=1\times 10^{-5}

The ratio of hydrogen ions in solution of pH equal to 11 to the solution of pH equal to 6 is 1\times 10^{-5}.

Difference between the H^+ ions at both pH:

1\times 10^{-6} M-1\times 10^{-11} M=9.99\time 10^{-7} M

This means that Hydrogen ions in a solution at pH = 7 has 9.99\time 10^{-7} M ions fewer than in a solution at a pH = 6

6 0
3 years ago
Read 2 more answers
At 298 K, Kc = 1.45 for the following reaction 2 BrCl (g) Br2(g) + Cl2(g) A reaction mixture was prepared with the following ini
tester [92]

Answer:

[BrCl]=0.02934M

[Br2]=[Cl2]=0.03533M

Explanation:

First, consider the Kc definition:

1.5=K_{c}=\frac{[Br_{2}][Cl_{2}]}{[BrCl]^{2}}

It is necessary to define a new variable, 'x', as the amount of moles of Br2 that are produced. If 'x' moles of Br2 are produced, the moles of the compounds will be calculated as;

n_{Br_{2}}=n_{{Br_{2}}^{0}} +x\\n_{Cl_{2}}=n_{{Cl_{2}}^{0}} +x\\n_{BrCl}=n_{{BrCl}}^{0}} -2x\\

Where the zero superscript means the initial moles.

Dividing the last equations by the volume, which is constant because the reaction does not change the total moles number (2 moles of BrCl produce 2 moles, one of Cl2 and another of Br2), we have the molarity equations for all species:

M_{Br_{2}}=M_{{Br_{2}}^{0}} +x\\M_{Cl_{2}}=\\M_{BrCl}=M_{{BrCl}}^{0}} -2x\\

And now 'x' is a change in molarity.

Replacing these in the Kc equation we have:

1.45=\frac{({M_{{Br_{2}}}^{0}}+x)({M_{{Cl_{2}}}^{0}}+x)}{({M_{{BrCl}}^{0}}-2x)^{2}}

Where the only unknown is 'x'. So, let's solve the equation:

1.45=\frac{(0.03+x)(0.03+x)}{(0.04-2x)^{2} } \\1.45(0.04-2x)^2=(0.03+x)^2\\1.45(0.0016-0.16x+4x^2)=0.0009+0.06x+x^2\\4.8x^2-0.292x+0.00142=0\\x_{1}=0.0555\\x_{2}=0.00533

The result x_{1}=0.0555 lacks of sense because it will give a negative concentration for BrCl, so the result is x_{2}=0.00533<u>.</u>

Applying the result, the concentrations at equilibrium are:

M_{Br_{2}}=M_{Cl_{2}}=0.03+0.00533=0.03533M\\M_{BrCl}=0.04-2*0.00533=0.02934M

If you calculate Kc with this concentrations it will give 1.45 as a result.

Greets, I will be happy to solve any doubt you have.

8 0
4 years ago
buffer consists of 0.50 M NaH2PO4 and 0.40 M Na2HPO4. Phosphoric acid is a triprotic acid (K a 1 = 7.2 × 10 − 3, K a 2 = 6.3 × 1
Levart [38]

Answer:

The Ka value used for the buffer system is :- Ka2 = 6.3\times 10^{-8} pKa2 = 7.2

pH = 7.1

Explanation:

The buffer system given in the question is :-

NaH_2PO_4 and Na_2HPO_4

The reaction taking place is:-

H_2PO_4^-\rightleftharpoons HPO_4^{2-}

The Ka value used for the buffer system is :- Ka2 = 6.3\times 10^{-8} pKa2 = 7.2

The pH can be calculated as:-

pH=pKa+\log\frac{Na_2HPO_4}{NaH_2PO_4}

pH = 7.2 + log 0.40/0.50 = 7.1

7 0
4 years ago
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