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Stels [109]
3 years ago
7

What atomic or hybrid orbitals make up the sigma bond between ge and h in germanium hydride, geh4 ?

Chemistry
2 answers:
ozzi3 years ago
8 0
<span>Germane is the chemical compound with the formula GeH₄, and the germanium analogue of methane. It is the simplest germanium hydride and one of the most useful compounds of germanium.


</span>In chemistry, sigma bonds (σ bonds) are the strongest type of covalent chemical bond. They are formed by head-on overlapping between atomic orbitals. Sigma<span> bonding is most simply defined for diatomic molecules using the language and tools of symmetry groups.
</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
ludmilkaskok [199]3 years ago
6 0

Answer:

Electron geometry: tetrahedral, sp3 hybridized.

Explanation:

Hello,

Since four atoms of hydrogen are surrounding the germanium in this case, a sp3 hybrization is making up the sigma bons for this molecular compound. The germanium atom can use its singly occupied p-type orbitals, to form two covalent bonds with two hydrogen atoms, and use the s and the other p to form the other two covalent bonds with the remaining hydrogens.

Best regards.

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Herbie left his house at 3:30 p.m. and arrived at Kit’s house at 5:00 p.m. The two boys live 7 miles apart. What was Herbie’s av
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Answer:

Step-by-step explanation:

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Suppose they take t minutes to meet each other.

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A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83
guapka [62]

Answer:

a. 478.69 K

b. 939.43 cm^{3}

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

V_{o} = 785cm^{3} = 0.000785 m^{3}

T_{o} = 400K

P_{o} = 125 Kpa =  125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 m^{3}⋅Pa⋅K^{-1}⋅mol^{-1}

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5T_{1} - (34980 + 35.5T_{o})]

83.8 = (0.03 × 35.5) (T_{1} - 400K)

83.8 = 1.065 (T_{1}  - 400)

78.69 = (T_{1}  - 400)

T_{1} = 400 + 78.69

T_{1}  = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

V_{1}/T_{1} = V_{o}/T_{o}

V_{1}  =  V_{o}T_{1}/T_{o}

V_{1}   = (785 × 478.69)/400

V_{1}   = 939.43 cm^{3}

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P(V_{1}- V_{o}) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

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