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Basile [38]
3 years ago
9

In a water molecule (H2O), the oxygen atom has two unshared pairs of electrons and two bonding pairs of electrons. How do the un

shared pairs affect the bond angle?
Chemistry
1 answer:
vladimir1956 [14]3 years ago
4 0

Answer:

They reduce the bond angle to be slightly lower than the tetrahedral bond angle, approximately 104.45 degrees.

Explanation:

The unshared pair of electrons or lone pair electrons in order to have the minimum repulsion possible with each other pushes the other bonding pairs closer together making the bond angle smaller or bent.

The bond angle is slightly lower than the tetrahedral bond angle of 108 degrees, leaving the water molecule with a bent molecular geometry.

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Someone please answer the question attached.
nikdorinn [45]

Answer:

Option ( 1 )

Explanation:

This atom has a cub close pack structure, and therefore - the number of edge center present in X atoms =  4,

Number of unit cells present in X atoms = 6

____________________________________________________

Now the 6th coordination number of X atom = 6 * 4 = 24,

So respectively, the 3rd coordination number of X = 8 -

And thus the ratio between the 6th coordination number and the 3rd coordination number = 24 / 8 = 3,

Option ( 1 )

<u><em>Hope that helps!</em></u>

5 0
3 years ago
Trust me when I say this.
eimsori [14]

Answer:

thank you

So much this was very nice.

6 0
3 years ago
Read 2 more answers
3. How much energy is needed to raise 45 grams of water from 40°C to 115 °C?
Dafna1 [17]

Answer:

Q = 114349.5 J

Explanation:

Hello there!

In this case, since this a problem in which we need to calculate the total heat of the described process, it turns out convenient to calculate it in three steps; the first one, associated to the heating of the liquid water from 40 °C to 100 °C, next the vaporization of liquid water to steam at constant 100 °C and finally the heating of steam from 100 °C to 115 °C. In such a way, we calculate each heat as shown below:

Q_1=45g*4.18\frac{J}{g\°C}*(100\°C-40\°C)=11286J\\\\Q_2=45g* 2260 \frac{J}{g} =101700J\\\\Q_3=45*2.02\frac{J}{g\°C}*(115\°C-100\°C)=1363.5J

Thus, the total energy turns out to be:

Q_T=11286J+101700J+1363.5J\\\\Q_T=114349.5J

Best regards!

5 0
3 years ago
What is the mass in grams of 5.00moles of CH4?
anastassius [24]

Answer:

1. 80g

2. 1.188mole

Explanation:

1. We'll begin by obtaining the molar mass of CH4. This is illustrated below:

Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol

Number of mole of CH4 from the question = 5 moles

Mass of CH4 =?

Mass = number of mole x molar Mass

Mass of CH4 = 5 x 16

Mass of CH4 = 80g

2. Mass of O2 from the question = 38g

Molar Mass of O2 = 16x2 = 32g/mol

Number of mole O2 =?

Number of mole = Mass /Molar Mass

Number of mole of O2 = 38/32

Number of mole of O2 = 1.188mole

6 0
3 years ago
2. What is the volume of 40.0 g of N2 at STP?
vampirchik [111]

Answer:

Molar mass of

N

2

=

28 g/mol

Moles of

40.5 g N

2

gas

=

40.5 g

28 g/mol

=

1.45 mol

1 mol

of any gas at

STP

occupies

22.4 L

of volume.

1.45 mol

of

N

2

gas will occupy

1.45

mol

×

22.4 L

1

mol

=

32.48 L

3 0
2 years ago
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