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3 years ago

For each of the following compounds, identify what type of bonding holds them together.

Chemistry

1 answer:

svet-max [94.6K]3 years ago

4 0

Explanation

NaCl: Ionic crystal lattice forces

Hg: Metallic bonding

CO₂: London dispersion forces

CH₄: London dispersion forces

Li₂O: Ionic crystal lattice forces

Ag: Metallic bonds

Ionic crystal lattice forces are strong electrostatic force of attraction between oppositely charged ions arranged into a crystal lattice of ionic compound. NaCl and Li₂O are ionic compounds

London dispersion forces holds the molecules of carbon dioxide and methane. They are weak attractions found between non-polar (and polar) molecules.

Metallic bonds exists between Mercury and Gold atoms. This is due to sea of electrons present.

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2H2 + O2 → 2H2O

pantera1 [17]

Okay
Mr (H2O)= 18g
therefore moles of H2O
is 720.8/18= 40.04mol
the ratio of H2 to O2 to H2O is
2 : 1 : 2
so moles of H2 is same as H2O here
H2= 40.04moles

moles of O2 is half
so 40.04 x 0.5
20.02moles

grams of O2 is
its moles into Mr of O2
that's 20.02 x 32 = 640.64g

6 0

2 years ago

What is the mass of ether(0.71) which can be put into a beaker holding 130ml

UkoKoshka [18]

D=m/v ⇒ m=d*v

d=density
m=mass
v=volume

d(ether)=0.71 gr/cm³=0.71 gr/ ml
v=130 ml

m=d*v
m=0.71 gr/ml*(130 ml)=92.3 g

Solution: m=92.3 g

6 0

3 years ago

For the reaction 2NH3(g) + 2O2(g)N2O(g) + 3H2O(l) H° = -683.1 kJ and S° = -365.6 J/K The standard free energy change for the rea

BlackZzzverrR [31]

Answer:

$\Delta G^{0}$ = -457.9 kJ and reaction is product favored.

Explanation:

The given reaction is associated with 2 moles of $NH_{3}$

Standard free energy change of the reaction ( $\Delta G^{0}$ ) is given as:

$\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}$ , where T represents temperature in kelvin scale

So, $\Delta G^{0}=(-683.1\times 10^{3})J-(273K\times -365.6J/K)=-583291.2J$

So, for the reaction of 1.57 moles of $NH_{3}$ , $\Delta G^{0}=(\frac{1.57}{2})\times -583291.2J=-457883.592J=-457.9kJ$

As, $\Delta G^{0}$ is negative therefore reaction is product favored under standard condition.

6 0

3 years ago

A balloon contains 2500 ml of helium gas at 75 degrees Celsius. What is the final volume in millimeters of gas when the temperat

REY [17]

I believe the answer is 20 degrees Celsius

3 0

2 years ago

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands

Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

$E=\frac{R_y}{n^2}$

$R_y=2.18\times 10^{-18} J$ = Rydberg energy

n = principal quantum number of the orbital

Energy of 11th orbit = $E_{11}$

$E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J$

Energy of 10th orbit = $E_{10}$

$E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J$

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

$E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J$

$=E'=0.38\times 10^{-20} J$

$\lambda =\frac{hc}{E'}$ (Planck's' equation)

$\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}$

$\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m$

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

3 0

2 years ago