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3 years ago

why is it easier for water to flow through rock and sediment with larger pores than through rock and sediment with smaller pores

Chemistry

2 answers:

andrey2020 [161]3 years ago

6 0

Answer:

Because of its permeability.

Explanation:

It is on quizlet. I don't think it is right though, I am sorry.

joja [24]3 years ago

6 0

Answer:

Larger pores mean less friction between flowing water and the sides of the pores. Smaller pores mean more friction between flowing water and pore walls and more twists and turns for the water to navigate. A permeable material has a greater number of larger, well-connected pore spaces. An impermeable material has fewer, smaller pores that are poorly connected.

Explanation:

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What is the approximate pressure of a storage cylinder of recovered R-404A that does not contain any non-condensable impurities

gregori [183]

Answer:

288 psig

Explanation:

4 0

3 years ago

Read 2 more answers

how many calories is in one peanut if the volume of water is 10 mL and the water temperature is rise 4 degrees celsius

liraira [26]

Answer:

40.02 calories

Explanation:

V = 10 mL = 10g

we know t went up by 4°C, this is our ∆t as it is a change.

Formula that ties it together: Q = mc∆t

where,

Q = energy absorbed by water

m = mass of water

c = specific heat of water (constant)

∆t = temperature change

Q = (10 g) x (4.186 J/g•°C) x (4°C)

Q = 167.44 J

Joules to Calories:

167.44 J x 1 cal/4.184 J = 40.02 calories

(makes sense as in image it is close to the value).

4 0

3 years ago

For the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g)

Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

Explanation :

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = -45.6 kJ = -45600 J

$\Delta S^o$ = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

$\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)$

$\Delta G^o=-12666.6J=-12.7kJ$

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln k$

where,

$\Delta G^o$ = standard Gibbs free energy = -12666.6 J

R = gas constant = 8.314 J/K.mol

T = temperature = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k$

$k=3.35\times 10^{2}$

Therefore, the value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

3 0

3 years ago

Consider the reaction of Mg3N2 with H2O to form Mg(OH)2 and NH3. If 4.33 g H2O is reacted with excess Mg3N2 and 6.26 g of Mg(OH)

Len [333]

Answer:

89.34%

Explanation:

First, write a balanced reaction.

Mg3N2 + 6H2O --> 3Mg (OH)2 + 2NH3

Next determine the moles of the known substance, or limiting reagent ( H2O)

n= m/MM

n ( H2O) = 4.33/(1.008×2)+16

n(H2O)= 0.2403

Use the mole ratio to find the moles of Mg(OH)2

0.2403 ÷2

n (Mg (OH)2) = 0.1202

Next, find the theoretical mass of Mg (OH)2 that should have been produced

m= n × MM

m= 0.1202 × (24.305 + (16×2) +(1.008 ×2))

=7.007g

To find percentage yield, divide the experimental amount by the theoretical amount and multiply by 100.

6.26/ 7.007 × 100

=89.34%

7 0

3 years ago

Here's a one-batch sample of Just Lemons lemonade production. Determine the percent yield and amount of leftover ingredients for

Bumek [7]

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4 0

3 years ago