A water tank can hold 1 m3 of water. When it’s empty, how much liters is needed to refill it?

According to the following reaction, what volume of 0.244 M KCl solution is required to react exactly with 50.0 mL of 0.210 M Pb

svetoff [14.1K]

Answer:

86 mL

Explanation:

First find the moles of Pb (NO3)2

n=cv

where

c ( concentration)= 0.210 M

v ( volume in L) =0.05

n= 0.210 × 0.05

n= 0.0105

Using the mole ratio, we can find the moles of KCl by multiplying by 2

n (KCl) =0.0105 ×2

=0.021

v (KCl)= n/ c

= 0.021/ 0.244

=0.08606557377

=0.086 L

= 86 mL

8 0

2 years ago

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Hikers have noticed that a sealed bag of potato chips puffs up when taken to the top of a mountain. Suppose that at the valley f

Dmitriy789 [7]

<h3>Answer:</h3>

0.95 atm

<h3>Explanation:</h3>

We are given;

Initial pressure, P1 = 1.0 atm

Initial temperature, T1 =298 K (25°C + 273)

Initial volume, V1 = 0.985 L

Final temperature, T2 = 295 K (22°C + 273)

Final volume, V2 = 1.030 L

We are required to find final air pressure;

Using the combined gas law;

$\frac{P1V1}{T1} =\frac{P2V2}{T2}$

To get, P2 ;

$P2=\frac{P1V1T2}{T1V2}$

$P2=\frac{(1.0)(0.985L)(295)}{(1.030)(298)}$

$P2 = 0.947 atm$

= 0.95 atm

Therefore, the air pressure at the top of the mountain is 0.95 atm

3 0

2 years ago

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NEED HELP!!!

Anit [1.1K]

Answer:

energy is released in the reaction.

Explanation:

3 0

2 years ago

Calculate the amount of energy in kilojoules needed to change 459 g of water ice at −10 ∘C to steam at 125 ∘C. The following con

egoroff_w [7]

Answer: $1.41\times 10^3kJ$

Explanation:-

The conversions involved in this process are :

$(1):H_2O(s)(-10^0C)\rightarrow H_2O(s)(0^0C)\\\\(2):H_2O(s)(0^0C)\rightarrow H_2O(l)(0^0C)\\\\(3):H_2O(l)(0^0C)\rightarrow H_2O(l)(100^0C)\\\\(4):H_2O(l)(100^0C)\rightarrow H_2O(g)(100^0C)\\\\(5):H_2O(g)(100^0C)\rightarrow H_2O(g)(125^0C)$

Now we have to calculate the enthalpy change:

$\Delta H=[n\times c_{ice}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[n\times c_{water}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[n\times c_{steam}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change = ?

m = mass of water = 459 g

$c_{s}$ = specific heat of ice = $36.57J/mol^0C$

$c_{l}$ = specific heat of water = $75.40J/mol^0C$

$c_{g}$ = specific heat of steam = $36.04J/gmol^0C$

n = number of moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{459g}{18g/mole}=25.5mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

$\Delta H_{vap}$ = enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole

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

$\Delta H=[25.5mole\times 36.57J/mol^0C\times (0-(-10))^0C]+25.5mole\times 6010J/mole+[25.5mole\times 75.40J/mol^0C\times (100-0)^0C]+25.5mole\times 40670J/mole+[25.5mole\times 36.04J/gmol^0C\times (125-100)^0c]$