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3 years ago

What kind of bonding would exist between sulfur and hydrogen?

Chemistry

1 answer:

Alexandra [31]3 years ago

5 0

Answer:

nonpolar covalent

Explanation:

Comparing their electronegativities will help determine the type of bond. Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The difference in electronegativity (ΔEN) is used to determine bond type. Hydrogen has an electronegativity of 2.20 and sulfur has 2.58. The difference is 0.38,so the electrons are shared in a nonpolar covalent bond.

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High self-monitors prefer situations in which clear expectations exist regarding how they're supposed to communicate. True False

xeze [42]

Answer:

True

Explanation:

This are acts or actions that concur with situational expectations.

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3 years ago

What tools would you use to measure the density of a rectangular block?

dimulka [17.4K]

A scale and a ruler. The scale to measure the mass, and a ruler to measure the volume.

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3 years ago

In the Mond process for the purification of nickel, carbon monoxide is reacted with heated nickel to produce Ni(CO)4, which is a

igor_vitrenko [27]

<u>Answer:</u> The equilibrium constant for this reaction is $1.068\times 10^{6}$

<u>Explanation:</u>

The equation used to calculate standard Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}]$

For the given chemical reaction:

$Ni(s)+4CO(g)\rightleftharpoons Ni(CO)_4(g)$

The equation for the standard Gibbs free change of the above reaction is:

$\Delta G^o_{rxn}=[(1\times \Delta G^o_{(Ni(CO)_4(g))})]-[(1\times \Delta G^o_{(Ni(s))})+(4\times \Delta G^o_{(CO(g))})]$

We are given:

$\Delta G^o_{(Ni(CO)_4(g))}=-587.4kJ/mol\\\Delta G^o_{(Ni(s))}=0kJ/mol\\\Delta G^o_{(CO(g))}=-137.3kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(1\times (-587.4))]-[(1\times (0))+(4\times (-137.3))]\\\\\Delta G^o_{rxn}=-38.2kJ/mol$

To calculate the equilibrium constant (at 58°C) for given value of Gibbs free energy, we use the relation:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = Standard Gibbs free energy = -38.2 kJ/mol = -38200 J/mol (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = $58^oC=[273+58]K=331K$

$K_{eq}$ = equilibrium constant at 58°C = ?

Putting values in above equation, we get:

$-38200J/mol=-(8.314J/Kmol)\times 331K\times \ln K_{eq}\\\\K_{eq}=e^{13.881}=1.068\times 10^{6}$

Hence, the equilibrium constant for this reaction is $1.068\times 10^{6}$

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