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kow [346]
3 years ago
10

Calculate the enthalpy change, ΔH, for the process in which 31.6 g of water is converted from liquid at 3.2 ∘C to vapor at 25.0°

C . For water, ΔHvap = 44.0 kJ/mol at 25.0°C and Cs = 4.18 J/(g*°C) for H2O(l).
Chemistry
1 answer:
amm18123 years ago
5 0

Answer:

Enthalpy change = 74.36 kJ

Explanation:

Enthalpy change is defines as the heat absorbed or evolved during a chemical reaction at constant pressure and volume

Reaction:

H2O(l) --> H2O(g)

DHr = mCsDT

Where m is the mass of water

Cs is the specific heat capacity

DT is the temperature difference

= 31.6 * 4.18 * (25 - 3.2)

= 2879.518 J

DHvap = 44kJ/mol

Moles of water = mass/molecular weight

Molecular weight = (1*2) + 16

= 18g/mol

Moles of water = 31.6/18

= 1.756 moles

DHvap = 44 * 1.756

= 77.244kJ

DH = DHproduct - DHreactant

DHproduct = DHvap = 77.24kJ

DHreactant = DHr = 2879.518J = 2.880kJ

= 77.24 - 2.880

DH = 74.36kJ

Enthalpy change = 74.36 kJ

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The pH of a 1.0M solution of butanoic acid HC4H7O2 is measured to be 2.41. Calculate the acid dissociation constant Ka of butano
Lubov Fominskaja [6]

Answer:

Ka = 1.52 E-5

Explanation:

  • CH3-(CH2)2-COOH ↔ CH3(CH2)2COO-  + H3O+

⇒ Ka = [H3O+][CH3)CH2)2COO-] / [CH3(CH2)2COOH]

mass balance:

⇒<em> C</em> CH3(CH2)2COOH = [CH3(CH2)2COO-] + [CH3(CH2)2COOH] = 1.0 M

charge balance:

⇒ [H3O+] = [CH3(CH2)2COO-]

⇒ Ka = [H3O+]²/(1 - [H3O+])

∴ pH = 2.41 = - Log [H3O+]

⇒ [H3O+] = 3.89 E-3 M

⇒ Ka = (3.89 E-3)² / ( 1 - 3.89 E-3 )

⇒ Ka = 1.519 E-5

3 0
3 years ago
What is the balenced equation of Cu+H2SO4---&gt; CuSO4+H2O+SO2
algol [13]
Balanced chemical equation :

1 Cu + 2 H2SO4 = 1 CuSO4 + 2 H2O + 1 SO<span>2</span>

hope this helps!.

8 0
3 years ago
In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for
inessss [21]

Answer:

The standard cell potential of the reaction is 0.78 Volts.

Explanation:

Cu^{2+}(aq)+Fe(s)\rightarrow Cu(s)+Fe^{2+}(aq)

Reduction at cathode :

Cu^+(aq)+2e^-\rightarrow Cu(s)

Reduction potential of  Cu^{2+} to Cu=E^o_{1}=0.34 V

Oxidation at anode:

Fe(s)\rightarrow Fe^{2+}(aq)+2e^-

Reduction potential of  Fe^{2+} to Fe=E^o_{2}=-0.44 V

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}

Putting values in above equation, we get:

E^o_{cell}=0.34V -(-0.44 V)=0.78 V

The standard cell potential of the reaction is 0.78 Volts.

3 0
3 years ago
Fred mixes a solution using two types of solutions: 0.60 liters containing 10% alcohol and 0.40 liters containing 35% alcohol. W
insens350 [35]

The alcohol concentration of the mixed solution is 20%

Simplification :

Based on the given condition, formulate :

35% ×0.40 + 0.6 ×10% ÷{ 0.4+0.6}

Calculate the product :\frac{0.14+0.06}{0.4 + 0.6}

Calculate the sum or difference : \frac{0.2}{1}

Any fraction with denominator 1 is equal to numerator : 0.2

Multiply a number to both numerator, denominator : 0.2 ×\frac{100}{100}

Calculate the product or quotient : \frac{20}{100}

A fraction with denominator equals to 100 to a percentage 20%.

How do you find the concentration of a mixed solution?

In general when your are mixing two different concentrations together first calculate number of moles for each solution (n=CV ,V-in liter) then add them together it will be total moles,then concentration of mixture will be = total moles / total volume(liter).

Learn more about concentration of alcohol :

brainly.com/question/13220698

#SPJ4

5 0
1 year ago
Pennies are made of zinc coated with copper. Copper has a work function of 4.7 eV. The ozone layer blocks nearly all solar radia
Vadim26 [7]

Answer:

yes  it can  λ =265  nm

Explanation:

Here we will use  the relationship

   E = h c/λ  ∴   λ  = E/ hc  where

                                          h= Plank's constant

                                           c= Speed of light

                                           λ = Wavelength = ?

Substituting

note need E in J ,

E = 4.7 eV x 1.602 x 10⁻¹⁹ J/eV = 7.5 x 10⁻¹⁹ J)

λ = 7.5 x 10 ⁻¹⁹ J / (  6.626 x 10⁻³⁴ Js x 3 x 10^8) =  2.65 x 10⁻⁷ m = 2.65

=  2.65 x 10⁻⁷ m x 1 x 10⁹ nm/m = 265 nm

4 0
3 years ago
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