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3 years ago

12

Lee is in customer service. How can she best support her internal customer, Dan? O al Prioritize other tasks and customers O b)

Ask him a lot of personal questions O c) Give him her full attention when he's speaking O d) Check with him before completing basic tasks

1 answer:

Thepotemich [5.8K]3 years ago

7 0

Answer:

do i care lol LOOK AT THIS

VVVVVVVVVVVVVVVVVVV

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Octane (C8H18)reacts oxygen gas (O2) to form carbon dioxide and water

iren [92.7K]

Answer:

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Explanation:

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3 years ago

Air trapped in a cylinder fitted with a piston occupies 145 mL at 1.08 atm

pav-90 [236]

Answer: 0.0014 atm

Explanation:

Given that,

Original pressure of air (P1) = 1.08 atm

Original volume of air (T1) = 145mL

[Convert 145mL to liters

If 1000mL = 1l

145mL = 145/1000 = 0.145L]

New volume of air (V2) = 111L

New pressure of air (P2) = ?

Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law

P1V1 = P2V2

1.08 atm x 0.145L = P2 x 111L

0.1566 atm•L = 111L•P2

Divide both sides by 111L

0.1566 atm•L/111L = 111L•P2/111L

0.0014 atm = P2

Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm

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3 years ago

question #5: corals are sensitive to temperature . if temperatures are raised , it will cause the coral to bleach . what does bl

larisa86 [58]

Answer:

coral bleaching is when the temperature of the water changes (even slight changes in temperature effect coral) and the coral reacts negatively to it causing it to die. unfortunately corals cannot recover once they have been bleached.

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3 years ago

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An atom can be compared to our solar system because: WILL MAKE BRAINLIEST!

Sindrei [870]

Answer:

uh i think its D All of the above

Explanation:

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3 years ago

It is found that a gas undergoes a first-order decomposition reaction. If the rate constant for this reaction is 8.1 x 10-2 /min

kap26 [50]

Answer:

28.43 min

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = $8.1\times 10^{-2}$ min⁻¹

Initial concentration $[A_0]$ = 0.1 M

Final concentration $[A_t]$ = $1.0\times 10^{-2}$ M

Time = ?

Applying in the above equation, we get that:-

$1.0\times 10^{-2}=0.1e^{-8.1\times 10^{-2}\times t}$

$0.1e^{-8.1\times \:10^{-2}t}=10^{-2}$

$e^{-8.1\times \:10^{-2}t}=\frac{1}{10}$

$\ln \left(e^{-8.1\times \:10^{-2}t}\right)=\ln \left(\frac{1}{10}\right)$

$t=28.43\ min$

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3 years ago