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4 years ago

Why is the time needed to vaporize the sample of water significantly greater than the time needed to melt the sample?

2 answers:

7 0

Significantly more energy is required to vaporize water because the inter molecular forces between molecules are being completely broken

sweet [91]4 years ago

4 0

Since the entomb sub-atomic powers must be totally broken overcome to end up a gas. The powers essentially should be relaxed when softening happens.

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For the reaction: CO(g) + H2O(g) â‡Œ CO2(g) + H2(g) the value of Kc is 1.845 at a specific temperature. We place 0.500mol CO and

patriot [66]

Answer:

Explanation:

In the equilibrium:

CO(g) + H₂O(g) ⇄ CO₂(g) + H₂(g)

Kc is:

Kc = 1.845 = [CO₂] [H₂] / [CO] [H₂O]

Where [] are equilibrium concentrations of each species

Initial concentrations:

[CO] = 0.500mol / 1.00L = 0.500M

[H₂O] = 0.500mol / 1.00L = 0.500M

In equilibrium, concentrations will be:

[CO] = 0.500M - X

[H₂O] = 0.500M - X

[CO₂] = X

[H₂] = X

Where X is reaction coordinate. The amount of reactant that reacts producing products.

Replacing in Kc expression:

1.845 = [CO₂] [H₂] / [CO] [H₂O]

1.845 = [X] [X] / [0.500M - X] [0.500M - X]

1.845 = X² / X² - X + 0.25

1.845X² - 1.845X + 0.46125 = X²

0.845X² - 1.845X + 0.46125 = 0

Solving for X:

X = 0.288M; Right solution

X = 1.9M; False solution: Produce negative concentrations.

Replacing, equilibrium concentrations are:

[CO] = 0.500M - X = 0.212M

[H₂O] = 0.500M - X = 0.212M

[CO₂] = X = 0.288M

[H₂] = X = 0.288M

3 0

3 years ago

The _ and _ are part of the digestive system. chose ONE from each of these list.

lina2011 [118]

Answer:

liver and esophagus.

Explanation:

Kidneys are part of the endocrine system, trachea and bronchi are part of the respiratory system

4 0

3 years ago

Read 2 more answers

what is the mass of carbon dioxide which contain the same number of molecules as are contained in 14 gram of oxygen?

Ratling [72]

Answer:

Mass of CO2 WILL BE ~ 9.33 g

Explanation:

Moles of O2 = 14/18

Let the mass of CO2 be x

Then moles of CO2 will be = x/12

moles of CO2 = moles of O2

x/12 = 14/16

x = 9.33 grams

3 0

3 years ago

Two Carnot engines are operated in series with the exhaust (heat output) of the first engine being the input of the second engin

OlgaM077 [116]

Answer:

(a) 140 F

(b) The temperature rise at the point where the heat is dumped is 2.51 degC

Explanation:

(a) Considering T1 the temperature of input of the first engine, T2 the temperature of the exhaust of the first engine (and input of the second engine) and T3 the exhaust of the second engine, if both engines have the same efficiency we have:

$\eta=1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}$

The temperatures have to be expressed in Rankine (or Kelvin) degrees

$1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}\\\\\frac{T_1}{T2}=\frac{T_2}{T_3}\\\\(T_2)^{2} =T_1*T_3\\\\T_2=\sqrt{T_1*T_3} =\sqrt{(459.67+260)*(459.67+40)}= \sqrt{719.67*499.67}\\\\ T_2=599 \, R= (599-459.67) ^{\circ} F=140^{\circ} F$

(b) The Carnot efficiency of the cycle is

$\eta_{c}=1-Th/Ts=1-(273+20)/(273+315)=0.502$

If the efficiency of the plant is 60% of the Carnot efficiency, we have

$\eta=0.6*\eta_{c}=0.6*0.502=0.302$

The heat used in the plant can be calculated as

$Q_i=W/\eta=750MW/0.302=2483MW$

And the heat removed to the heat sink is

$Q_o=Qi-W=2483-750=1733MW$

If the flow of the river is 165 m3/s, the heat per volume in the sink is

$\frac{Q_o}{f} =\frac{1733 MJ/s}{165 m3/s}= 10.5MJ/m3$

Considering a heat capacity of water C=4.1796 kJ/(kg*K) and a density ρ of 1000 kg/m3, the temperature rise of the water is

$\Delta Q=C*\Delta T\\\Delta T=(1/C)*\Delta Q\\\Delta T=(\frac{1}{4.1796\frac{kJ}{kgK} } )*10,500\frac{kJ}{m3}*\frac{1m3}{1000kg}\\\Delta T= 2.51 ^{\circ}C$

8 0

4 years ago

1.How many moles of each element are in 0.0250 mol of K 2 Cr O 4? 2. How many moles of ammonium ions are in 4.50 mol of (N H 4 )

noname [10]

Answer:

1. 0.0500 moles K, 0.0250 moles Cr and 0.1000 moles O.

2. 9.00 moles ammonium ions are present.

Explanation:

1. In 1mol of K2CrO4 there are 2 moles of K, 1 mole of Cr and 4 moles of O.

The moles in 0.0250 moles K2CrO4 are:

0.0250mol * 2 = 0.0500 moles K

0.0250mol * 1 = 0.0250 moles Cr

0.0250mol * 4 = 0.1000 moles O

2. In 1 mole of (NH4)2CO3 there are 2 moles of ammonium ions, NH₄⁺.

In 4,50 moles are:

4.50 moles * 2 = 9.00 moles ammonium ions are present

5 0

3 years ago