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tigry1 [53]
3 years ago
10

Temperature of 50g of methanol from 20 degrees Celsius to 70 degrees Celsius

Chemistry
1 answer:
olganol [36]3 years ago
4 0

[ Q ] = [ Delta U ] + [ W sub B ]

No work is done. -----> [ W sub B ] = 0.0

[ Delta U ] = [ m ] [ C sub V ] [ T2 - T1 ]

[ Delta U ] = [ 50 ] [ 0.58 ][70 - 20 ]

[ Delta U ] = 58 cal

Q = [ Delta U ] = 58cal
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Relative atomic mass or atomic weight is a dimensionless physical quantity defined as the ratio of the average mass of atoms of a chemical element in a given sample to the atomic mass constant. The atomic mass constant is defined as being 1/12 of the mass of a carbon-12 atom.

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3 years ago
Electrolysis is an endothermic process when an electric current is passed through water and separates it into hydrogen and oxyge
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3 years ago
How much energy is required to raise the temperature of 10.7 grams of gaseous helium from 22.1 °C to 39.4 °C ?
Rainbow [258]

Answer:

Q = 2640.96 J

Explanation:

Given data:

Mass of He gas = 10.7 g

Initial temperature = 22.1°C

Final temperature = 39.4°C

Heat absorbed = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree. Specific heat capacity of He is 14.267 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 39.4°C - 22.1°C

ΔT = 17.3°C

Q = 10.7 g× 14.267 J/g.°C ×  17.3°C

Q = 2640.96 J

5 0
3 years ago
Which choice gives the correct oxidation numbers for all three elements in rb2so3 in the order that the elements are shown in th
vlabodo [156]
<h3>Answer:</h3>

             Rb  =  + 1

              S    =  + 4

              O   =  - 2

<h3>Explanation:</h3>

                   Oxidation states of the elements were calculated keeping in mind the basic rules of assigning oxidation states which included assignment of +1 charge to first group elements i.e. Rubidium (Rb) and assignment of -2 charge to Oxygen atom. Then the oxidation state of Sulfur was calculated as follow,

Rb₂ + S + O₃  =  0

Above zero (0) means that the overall molecule is neutral.

Putting values of Rb and O,

                                           (+1)₂ + S + (-2)₃  =  0

                                           (+2) + S + (-6)  =  0

                                           +2 + S - 6  = 0

                                           S - 6  =  -2

                                          S  =  -2 + 6

                                           S  =  + 4

7 0
3 years ago
What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )
yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution :  Given,

Concentration (c) = 0.150 M

Acid dissociation constant = k_a=4.5\times 10^{-4}

The equilibrium reaction for dissociation of HNO_2 (weak acid) is,

                           HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+

initially conc.         c                       0         0

At eqm.              c(1-\alpha)                c\alpha        c\alpha

First we have to calculate the concentration of value of dissociation constant (\alpha ).

Formula used :

k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}

Now put all the given values in this formula ,we get the value of dissociation constant (\alpha ).

4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}

4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2

0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0

By solving the terms, we get

\alpha=0.0533

No we have to calculate the concentration of hydronium ion or hydrogen ion.

[H^+]=c\alpha=0.150\times 0.0533=0.007995 M

Now we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.007995 M)

pH=2.097\approx 2.1

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0
3 years ago
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