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2 years ago

What is a hot spot? Own words

1 answer:

6 0

Answer: Hotspot: A hotspot is a physical location where people can access the Internet, typically using Wi-Fi, via a wireless local area network (WLAN) with a router connected to an Internet service provider. ... While many public hotspots offer free wireless access on an open network, others require payment.

A hot spot is an area on Earth over a mantle plume or an area under the rocky outer layer of Earth, called the crust, where magma is hotter than surrounding magma. The magma plume causes melting and thinning of the rocky crust and widespread volcanic activity.

Explanation: MY own word's.

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1 year ago

How many atoms are in 3 moles of carbon?

Leokris [45]

Answer:

18.06 x 10^23

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2 years ago

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Heat energy is transferred by conduction whenever molecules?

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I think the answer is D

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2 years ago

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A rigid container of gas has a pressure of 1.72 atm and temperature of 21 oC. If the temperature increases to 85 oC, what is the

Vinil7 [7]

Answer:

the new pressure is 2.09 atm

Explanation:

you have to use gay lussac's law so the formula is

p1/t1 = p2/t2

and convert C to Kelvin k=C+273.15

1.72atm/294.15 = p2/358.15

solve for p2 by multiplying 358.15 on both sides

p2=2.09 atm

5 0

2 years ago

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m

Bond [772]

Answer:

$m_{H_2O}=4.86gH_2O$

Explanation:

Hello,

In this case, the described chemical reaction is:

$C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O$

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

$n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6$

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

$n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6$

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

$m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O$

Best regards.

3 0

2 years ago