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Angelina_Jolie [31]
3 years ago
6

An empty water bottle is full of air at 15°C and standard pressure. The volume of the bottle is 0.500 liter. How many moles of a

ir are in the bottle?
Chemistry
1 answer:
lord [1]3 years ago
3 0

Answer:  0.02 moles

Explanation: Using ideal gas equation:  

PV=nRT

n=\frac{PV}{RT}

P = pressure = standard pressure= 1 atm

V = volume = 0.500L

n = no of moles = ?

R = gas constant =0.0821 Latm\molK

T= temperature = 15^0C = (15+273)K= 288K

n=\frac{1\times 0.500}{0.0821\times 288}=0.02moles




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calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w
kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is -22Jmole^-^1k^-^1

(b) The normal boiling point of water (J·K−1·mol−1) is -109Jmole^-^1K^-^1

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is  109J/mole

Explanation:

Lets calculate

(a) - General equation -

      (\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p = -5_m(\beta )+5_m(\alpha ) =  -\frac{\Delta H}{T}

 \alpha ,\beta → phases

ΔH → enthalpy of transition

T → temperature transition

 (\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p == -\frac{\Delta_fH}{T_f}

            = \frac{-6.008kJ/mole}{273.15K} ( \Delta_fH is the enthalpy of fusion of water)

           = -22Jmole^-^1k^-^1

(b) (\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}

                                  = \frac{40.656kJ/mole}{373.15K} (\Delta_v_a_p_o_u_rH is the enthalpy of vaporization)

                               = -109Jmole^-^1K^-^1

(c) \Delta\mu =\Delta\mu(l)-\Delta\mu(s) =-S_m\DeltaT

[\mu(l-5°C)-\mu(l,0°C)] =  [\mu(s-5°C)-\mu(s,0°C)]=-S_mΔT

\mu(l,-5°C)-\mu(s,-5°C)=-Sm\DeltaT [\mu(l,0

\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)

     = 109J/mole

6 0
2 years ago
Given the reaction: solid sodium reacts with chlorine gas to form solid sodium chloride How many moles of the product result fro
lyudmila [28]

Answer:

number of moles of NaCl produce = 12 mol

Explanation:

Firstly, we need to write the chemical equation of the reaction and balance it .

Na(s) + Cl2(g) → NaCl(s)

The balanced equation is as follows:

2Na(s) + Cl2(g) → 2NaCl(s)

1 mole(71 g) of chlorine produces 2 moles(117  g) of sodium chloride

6 mole of chlorine gas will produce ? mole of sodium chloride

cross multiply

number of moles of NaCl produce = 6 × 2

number of moles of NaCl produce = 12 moles

number of moles of NaCl produce = 12 mol

3 0
3 years ago
What direction do the plates have to move in order for an earthquake to occur?
Bess [88]
they would be pushing together
4 0
3 years ago
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PLEASE HELP ME I NEED HELP
Crazy boy [7]

Answer:

<u>5 moles S x (36.02 g S/mole S) = 180.1 grams of S</u>

Explanation:

The periodic table has mass units for every element that can be correlated with the number of atoms of that element.  The relationship is known as Avogadro's Number.  This number, 6.02x10^{23} , is nicknamed the mole, which scientists found to be a lot more catchy, and easier to write than  6.02x10^{23}.  <u>The mole is correlated to the atomic mass of that element.</u>  The atomic mass of sulfur, S, is 36.02 AMU, atomic mass units.  <u>But it can also be read as 36.02 grams/mole.</u>

<u></u>

<u>This means that 36.02 grams of S contains 1 mole (6.02x</u>10^{23}<u>) of S atoms</u>.

<u></u>

This relationship holds for all the elements.  Zinc, Zn, has an atomic mass of 65.38 AMU, so it has a "molar mass" of 65.38 grams/mole.  ^5.38 grams of Zn contains 1 mole of Zn atoms.  

And so on.

5.0 moles of Sulfur would therefore contain:

(5.0 moles S)*(36.02 grams/mole S) = <u>180.1 grams of S</u>

Note how the units cancel to leaves just grams.  The units are extremely helpful in mole calculations to insure the correct mathematical operation is done.  To find the number of moles in 70 g of S, for example, we would write:

(70g S)/(36.02 grams S/mole S) = 1.94 moles of S.  [<u>Note how the units cancel to leave just moles</u>]

4 0
2 years ago
How many grams of lead would be required to make 119 grams of water
Ludmilka [50]

Answer:

(119 g H2O) / (18.01532 g H2O/mol) x (1 mol Pb / 2 mol H2O) x (207.21 g Pb/mol) = 684 g Pb

Explanation:

8 0
3 years ago
Read 2 more answers
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