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Nonamiya [84]
3 years ago
13

A 19-g piece of metal absorbs 186.75 joules of heat energy, and its temperature changes from 35°C to 175°C. Calculate the specif

i heat capacity of the metal
Chemistry
1 answer:
slavikrds [6]3 years ago
6 0

Answer:

0.0702J/g°C the specific heat capacity of the metal.

Explanation:m

Q=m\times c\times \Delta T=m\times c\times (T_{2}-T_{1})

where,

Q = heat absorbed by metal = 186.75 J

m_1 = Mass of metal= 19 g

T_1 = Initial  temperature of metal = 35^oC

T_2 =Final  temperature of metal = 175^oC

c = specific heat of metal= ?

186.75 J=19 g\times c\times (175^oC-35^oC)

c=\frac{186.75 J}{19 g\times (175^oC-35^oC)}

c=0.0702J/g^oC

0.0702J/g°C the specific heat capacity of the metal.

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What is the molarity of solution obtained when 5.71 g of sodium carbonate-10-water is dissolved in water and made up to 250.0 cm
disa [49]

We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.

The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O)  is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³

The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.

Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.

5.71 grams of Na₂CO₃.10 H₂O is equal to \frac{5.71}{286}= 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.

Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is \frac{0.0199 X 1000}{250} = 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³


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3 years ago
Which of the following statements is true? A) This reaction will be spontaneous only at high temperatures. B) This reaction will
viktelen [127]

Answer:

D) This reaction will be nonspontaneous only at high temperatures.

Explanation:

According the equation of Gibb's free energy -

∆G = ∆H -T∆S

∆G = is the change in gibb's free energy

∆H = is the change in enthalpy

T = temperature

∆S = is the change in entropy .

And , the sign of the  ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,

i.e. ,

if

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The reaction has the value for ∆H = negative , and ∆S = negative ,

Now ,

∆G = ∆H -T∆S

     = ( - ∆H ) - T( - ∆S )

     =  ( - ∆H ) +T(  ∆S )

Now, for making the reaction Spontaneous ΔG = negative ,

Hence ,

The temperature is low, then the value for  ΔG will be negative , i.e. , Spontaneous reaction .

And , vice versa , at higher temperature , the reaction will have ΔG positive , and the reaction will be non -Spontaneous reaction .

The standard free energy of formation will be zero , only for the compounds that are in their pure form ,

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In an undisturbed sequence of layers of rocks, the younger layers lie on top of the older layers
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I don't need to know why i just want the awnser
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Answer:

I remember doing this in 7th,

1. D

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I am in desperate need of help will give 40 points to whoever helps me!!!!! I have a quiz tomorrow but I have no idea how to do
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All of the questions here are pertaining to the colligative properties of a solution and the preparation of solutions. Maybe, it would be best if you understand the equations to be used in order to answer these questions.<span>
Freezing point depression or Boiling point elevation:

</span><span>ΔT = -K (m) (i)

</span>ΔT is the change in the freezing point or the boiling point not the freezing point/boiling point. Therefore, it should be added to the original value of the property of the solvent. 
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K is a constant called the molal freezing point depression constant and for the boiling point is the boiling point elevation constant. It is a property of the solvent. 
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m is the concentration of the solute in the solvent in terms of molality or kg solute/kg solvent. 
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i is the vant hoff factor which will represent the number of ions which the solute dissociates when in solution.</span>
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