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3 years ago

Explain why a Xenon atom is larger than a Krypton atom.

Chemistry

1 answer:

vovangra [49]3 years ago

3 0

It has more electrons, meaning more orbitals, pls give brainliest.

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What is the most important factor in determining climates.

Ira Lisetskai [31]

Latitude - Many factors influence the climate of a region. The most important factor is latitude because different latitudes receive different amounts of solar radiation. The maximum annual temperature of the Earth, showing a roughly gradual temperature gradient from the low to the high latitudes.

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3 years ago

How many moles of ammonia are produced when 5.0 moles of hydrogen react with excess nitrogen? Use this balanced equation for the

wel

3 mol H₂ → 2 mol NH₃

5 mol H₂ → x mol NH₃

x=2*5.0/3=3.3

n(NH₃)=3.3 mol

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3 years ago

What is the value of quantum numbers (n, l, ml, ms) of 2p^3?

eduard

Answer:

Explanation:

First digit of the 2p^3 gives you value of n, in this case its = 2, So, n= 2

Second alphabet gives you the value of l,

l=0 =s

l=1 =p

l=3=d

l=4=f

since "p" is the alphabet in 2p^3, so in your case lt shoudlbe = 1 right?

ml= -l to +l , that is -1, 0, +1

Ms= +1/2 or -1/2 alaways remains same foe evrything.

6 0

2 years ago

10 pointsss Where are tsunamis likely to originate?

ycow [4]

Tsunamis are likely to originate in the pacific ocean

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2 years ago

Read 2 more answers

The formation of ammonia is represented by the equation N2(g) + 3H2(g) ⇌ 2NH3(g). Determine the enthalpy of formation of ammonia

Savatey [412]

Answer:

$\Delta _fH_{NH_3}=-66\frac{kJ}{mol}$

Explanation:

Hello!

In this case, since the study of the bond energy allows us to compute the enthalpies of some reactions, for this combination reaction by which ammonia is yielded, we understand the enthalpy of reaction equals the enthalpy of formation of ammonia, and, in terms of the bonds energy we can write:

$\Delta _fH_{NH_3}=Delta _rH=\Sigma \Delta H(bonds \ broken)-\Sigma \Delta H(bonds \ formed)$

Whereas the bonds enthalpy of those bonds that get broken cover the N≡N and the three H-H bonds at the reactants side and the enthalpy of those bonds that are formed cover the six N-H bonds at the products; which means we obtain:

$\Delta _fH_{NH_3}=942\frac{kJ}{mol} +3*436\frac{kJ}{mol}-6*386\frac{kJ}{mol}\\\\\Delta _fH_{NH_3}=-66\frac{kJ}{mol}$

Which differs from the theoretical value that is -46 kJ/mol.

Best regards!

7 0

3 years ago