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Sonbull [250]
3 years ago
7

Waves crash on a beach one after another. Why doesn't water pile up on the beach

Chemistry
2 answers:
Klio2033 [76]3 years ago
6 0

sand

Explanation:

it is absorbing it

UNO [17]3 years ago
3 0

Answer:

Sand absorbers the water

Explanation:

On other days, however, when winds blow surface water toward land, the water can't just pile up on the beach, so water near the bottom has to compensate, moving offshore. ... One measures the speed and direction of flowing water by reflecting sound off particles carried by the currents.

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Mixtures of benzene and cyclohexane exhibit ideal behavior. A solution was created containing 1.5 moles of liquid benzene and 2.
goblinko [34]

Answer:

Vapour pressure of cyclohexane at 50°C is 490torr

Vapour pressure of benzene at 50°C is 90torr

Explanation:

Using Raoult's law, pressure of a solution is defined by the sum of the product sbetween mole fraction of both solvents and pressure of pure solvents.

P_{solution} = X_{A}P^0_{A}+X_{B}P^0_{B}

In the first solution:

X_{cyclohexane}=\frac{2.5mol}{2.5mol+1.5mol} =0.625

X_{benzene}=\frac{1.5mol}{2.5mol+1.5mol} =0.375

340torr = 0.625P^0_{A}+0.375P^0_{B} <em>(1)</em>

For the second equation:

X_{cyclohexane}=\frac{3.5mol}{3.5mol+1.5mol} =0.700

X_{benzene}=\frac{1.5mol}{3.5mol+1.5mol} =0.300

370torr = 0.700P^0_{A}+0.300P^0_{B}<em>(2)</em>

Replacing (2) in (1):

340torr = 0.625P^0_{A}+0.375(1233.3-2.333P^0_{A})

340torr = 0.625P^0_{A}+462.5-0.875P^0_{A}

-122.5torr = -0.250P°A

P^0_{A} = 490 torr

<em>Vapour pressure of cyclohexane at 50°C is 490torr</em>

And for benzene:

370torr = 0.700*490torr+0.300P^0_{B}

P^0_{B}=90torr

<em>Vapour pressure of benzene at 50°C is 90torr</em>

3 0
3 years ago
Sometimes a cloud may form at or near the ground when the ground cools at night. This cloud is known as...
melomori [17]

Answer:

im belive the answers fog

8 0
2 years ago
Read 2 more answers
A main group metal was studied and found to exhibit the following properties:
Alecsey [184]

Answer: we learned this not to long ago i think its a,c

Explanation:

6 0
3 years ago
How many moles of oxygen are required to produce 37.15 g CO2? 37.15 g CO2 = mol O2
NNADVOKAT [17]

Answer:

0.84 moles of oxygen are required.

Explanation:

Given data:

Mass of CO₂ produced = 37.15 g

Number of moles of oxygen = ?

Solution:

Chemical equation:

C + O₂     →     CO₂

Number of moles of  CO₂:

Number of moles = mass/molar mass

Number of moles = 37.15 g/ 44 g/mol

Number of moles = 0.84 mol

Now we will compare the moles of oxygen and carbon dioxide.

                          CO₂         :       O₂  

                              1           :         1

                            0.84       :       0.84

0.84 moles of oxygen are required.

6 0
3 years ago
Read 2 more answers
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
3 years ago
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