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Luba_88 [7]
3 years ago
10

A molecule has two atoms that are bonded to the central atom and one lone pair of electrons around this central atom. What is th

e shape of the molecule?
bent
linear
trigonal planar
trigonal pyramidal
Chemistry
2 answers:
alex41 [277]3 years ago
3 0
The shape of the molecule is trigonal planar.
For example, H₂O:
     ..
    O
  /    \
H     H

Sholpan [36]3 years ago
3 0

Answer:

bent

Explanation:

Let’s take the example of SO₂, here the central S atom has 6 valence electrons. S uses 1 pair of electron to form a double bond with each O atom. So, S in total uses four of its valence electrons to form two double bonds with two O atoms. Now after bond formation two electrons are still left on S, thus S has a lone pair of electron. In the absence of a lone pair of electron on S, SO₂ would have a linear structure, O=S=O. However, the lone pair-bond pair repulsion causes SO₂ to have a bent structure.

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Answer:

B. Concept Formation

Explanation:

All the other answer choices are not equivalent.

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3 years ago
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What is the volume of 0.80 grams of o2 gas at stp? (5 points) group of answer choices 0.59 liters 0.56 liters 0.50 liters 0.47 l
Vladimir [108]

Answer:

0.56L

Explanation:

This question requires the Ideal Gas Law:  PV=nRT where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.

Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units:R=0.0821\frac{L\cdot atm}{mol\cdot K}

Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means T=273.15K and P=1atm

Lastly, we must calculate the number of moles of O_2(g) there are.  Given 0.80g of O_2(g), we will need to convert with the molar mass of O_2(g).  Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2:  32g\text{ }O_2=1mol\text{ }O_2

Thus, \frac{0.80g \text{ }O_2}{1} \frac{1mol\text{ }O_2}{32g\text{ }O_2}=0.25mol\text{ }O_2=n

Isolating V in the Ideal Gas Law:

PV=nRT

V=\frac{nRT}{P}

...substituting the known values, and simplifying...

V=\frac{(0.025 mol \text{ }O_2)(0.0821\frac{L\cdot atm}{mol \cdot K} )(273.15K)}{(1atm)}

V=0.56L \text{ } O_2

So, 0.80g of O_2(g) would occupy 0.56L at STP.

5 0
2 years ago
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Answer:

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M2 = 2.5 M

V2 = ?

Solve for V2 --> V2 = M1V1/M2

V2 = (0.075 M)(200 L) / (2.5 M) = 6.0 L

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Answer:

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