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nikitadnepr [17]
3 years ago
13

In diamond, carbon atoms are bonded together by extended covalent bonds. what type of solid is diamond?

Chemistry
1 answer:
irinina [24]3 years ago
4 0

Answer: -

D. Network

Explanation: -

Diamond is an allotrope of carbon. In diamond each carbon atom makes four bonds to other carbon atoms.

They exist in tetrahedral shape.

Diamond has strong covalent bonds. They extend in all the three dimensions

Such covalent bonds are called network covalent bonds. They require significant amounts of energy to break.

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At 35 C, a sample of gas has a volume of 256 ml and a pressure of 720.torr. What would the volume
Natalija [7]

Answer: Volume would be 196.15 mL if the temperature were changed to 22^{o}C and the pressure to 1.25 atmospheres.

Explanation:

Given: T_{1} = 35^{o}C = (35 + 273) K = 308 K,     V_{1} = 256 mL,    

P_{1} = 720 torr (1 torr = 0.00131579 atm) = 0.947368 atm

T_{1} = 22^{o}C = (22 + 273) K = 295 K,       P_{2} = 1.25 atm  

Formula used to calculate volume is as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{1 atm \times 256 mL}{308 K} = \frac{1.25 atm \times V_{2}}{295 K}\\V_{2} = 196.15 mL

Thus, we can conclude that the volume would be 196.15 mL if the temperature were changed to 22^{o}C and the pressure to 1.25 atmospheres.

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In a certain city, electricity costs $0.15 per kW·h. What is the annual cost for electricity to power a lamp-post for 6.00 hours
Vanyuwa [196]

(a) Power of bulb is 100 W, converting this into kW.

1 W=\frac{1}{1000}kW

Thus,

100 W =\frac{100}{1000}kW=0.1 kW

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t=6\times 365=2190 h

Electricity used will depend on power and number of hours as follows:

E=P\times t=0.1 kW\times 2190 h=219 kW.h

The cost of electricity is $0.15 per kW.h thus, cost of electricity for 219 kW.h will be:

Cost=\$ (219\times 0.15)=\$ 32.85

Therefore, annual cost of incandescent light bulb is \$ 32.85

(b) Power of bulb is 25 W, converting this into kW.

1 W=\frac{1}{1000}kW

Thus,

100 W =\frac{25}{1000}kW=0.025 kW

The bulb is used for 6 hours per day for a year, in 1 year there are 365 days thus, total hours will be:

t=6\times 365=2190 h

Electricity used will depend on power and number of hours as follows:

E=P\times t=0.025 kW\times 2190 h=54.75 kW.h

The cost of electricity is $0.15 per kW.h thus, cost of electricity for 54.75 kW.h will be:

Cost=\$ (54.75\times 0.15)=\$ 8.21

Therefore, annual cost of fluorescent bulb is \$ 8.21.

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Answer:

Option c. Neutral

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