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3 years ago

6

The molar heat capacity of an unknown substance is 92.1 J/mol-K. If the unknown has a molar mass of 118 g/mol, what is the speci

fic heat (J/g-K) of this substance?

1 answer:

dlinn [17]3 years ago

8 0

Answer : The specific heat (J/g-K) of this substance is, 0.780 J/g.K

Explanation :

Molar heat capacity : It is defined as the amount of heat absorbed by one mole of a substance to raise its temperature by one degree Celsius.

1 mole of substance releases heat = 92.1 J/K

As we are given, molar mass of unknown substance is, 118 g/mol that means, the mass of 1 mole of substance is, 118 g.

As, 118 g of substance releases heat = 92.1 J/K

So, 1 g of substance releases heat = $\frac{92.1}{118}=0.780J/g.K$

Thus, the specific heat (J/g-K) of this substance is, 0.780 J/g.K

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With the kinetic energy of a sample increases the____ will also increase

klemol [59]

Answer:

Its Temperature.

Explanation:

Kinetic energy is the energy possessed by a body while in motion. Potential energy on the other hand is the energy possessed by a body based on its position.
Kinetic energy of a body is determined by the equation; K.E = 3RT/2;
From the equation, kinetic energy is directly proportional to the temperature of the system and therefore an increase in the kinetic energy of a sample causes an increase in temperature of the system.

4 0

2 years ago

A container of 56mL is at STP, The pressure is increased to 3.6atm and the volume to 162 mL.

Sindrei [870]

Answer:

T2 = 2843.1 oK. This is a huge temperature. Check it for errors.

Explanation:

Remark

This is the same question as the other one I've answered. Only the numbers have been altered.

Givens

v1 = 56 mL

P1 = 1 atm

T1 = 273o K

v2 = 162

P2 = 3.6 atm

T2 = ?

Formula

Vi * P1 / T1 = V2 * P2/T2

Solution

Rearrange the formula so T2 is on the left

T2 = V2 P2 * T1 / (V1 * P1) Now just put the numbers in.

T2 = 162 * 3.6* 273 / (56 *1)

T2 = 159213.6/56

T2 = 2843.1

3 0

2 years ago

I need answers plz<br>btw D says ice crystals

tangare [24]

Answer:

B. Hydrogen and oxygen molecules

General Formulas and Concepts:

<u>Chemistry - Reactions</u>

Reactants are always on the left side of the arrow
Products are always on the right side of the arrow

Explanation:

<u>Step 1: Define</u>

Reaction RxN: 2H₂ + O₂ → 2H₂O

<u>Step 2: Identify</u>

Reactants: H₂ and O₂

Products: H₂O

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2 years ago

Use the data given below to construct a Born-Haber cycle to determine the second ionization energy of Ca. Δ H°(kJ) Ca(s)→Ca(g) 1

Drupady [299]

Answer : The value of second ionization energy of Ca is 1010 kJ.

Explanation :

The formation of calcium oxide is,

$Ca(s)+\frac{1}{2}O_2(g)\overset{\Delta H_f}\rightarrow CaO(s)$

$\Delta H_f^o$ = enthalpy of formation of calcium oxide = -635 kJ

The steps involved in the born-Haber cycle for the formation of $CaO$ :

(1) Conversion of solid calcium into gaseous calcium atoms.

$Ca(s)\overset{\Delta H_s}\rightarrow Ca(g)$

$\Delta H_s$ = sublimation energy of calcium = 193 kJ

(2) Conversion of gaseous calcium atoms into gaseous calcium ions.

$Ca(g)\overset{\Delta H_I_1}\rightarrow Ca^{+1}(g)$

$\Delta H_I_1$ = first ionization energy of calcium = 590 kJ

(3) Conversion of gaseous calcium ion into gaseous calcium ions.

$Ca^{+1}(g)\overset{\Delta H_I_2}\rightarrow Ca^{+2}(g)$

$\Delta H_I_2$ = second ionization energy of calcium = ?

(4) Conversion of molecular gaseous oxygen into gaseous oxygen atoms.

$O_2(g)\overset{\Delta H_D}\rightarrow OI(g)$

$\frac{1}{2}O_2(g)\overset{\Delta H_D}\rightarrow O(g)$

$\Delta H_D$ = dissociation energy of oxygen = $\frac{498}{2}=249kJ$

(5) Conversion of gaseous oxygen atoms into gaseous oxygen ions.

$O(g)\overset{\Delta H_E_1}\rightarrow O^-(g)$

$\Delta H_E_1$ = first electron affinity energy of oxygen = -141 kJ

(6) Conversion of gaseous oxygen ion into gaseous oxygen ions.

$O^-(g)\overset{\Delta H_E_2}\rightarrow O^{2-}(g)$

$\Delta H_E_2$ = second electron affinity energy of oxygen = 878 kJ

(7) Conversion of gaseous cations and gaseous anion into solid calcium oxide.

$Ca^{2+}(g)+O^{2-}(g)\overset{\Delta H_L}\rightarrow CaO(s)$

$\Delta H_L$ = lattice energy of calcium oxide = -3414 kJ

To calculate the overall energy from the born-Haber cycle, the equation used will be:

$\Delta H_f^o=\Delta H_s+\Delta H_I_1+\Delta H_I_2+\Delta H_D+\Delta H_E_1+\Delta H_E_2+\Delta H_L$

Now put all the given values in this equation, we get:

$-635kJ=193kJ+590kJ+\Delta H_I_2+249kJ+(-141kJ)+878kJ+(-3414kJ)$

$\Delta H_I_2=1010kJ$

Therefore, the value of second ionization energy of Ca is 1010 kJ.

6 0

2 years ago

A sample of gas contains 0.1100 mol of HBr(g) and 5.00x10^-2 mol of Cl2(g) and occupies a volume of 5.85 L. The following reacti

Vesna [10]

The volume of the sample after the reaction : V₂=5.484 L

<h3>Further explanation</h3>

Given

Reaction

2HBr(g) + Cl2(g) => 2HCl (g)+Br2(g)

0.1100 mol of HBr(g)

5x10⁻² mol of Cl2(g)

Required

The volume of the sample after reaction

Solution

Find limiting reactant :

0.11/2 : 0.05/1 = 0.055 : 0.05

Limiting reactant : Cl₂

mol Products based on mol Cl₂

mol products = mol HCl + mol Br₂

mol products = 2/1x0.05 + 1/1x0.05

mol products = 3 x 0.05 = 0.15

mol reactants = 0.11 + 0.05 = 0.16

<em>From Avogadro's law, </em>

In the same T, P, and V, the gas contains the same number of molecules

So the ratio of gas volume will be equal to the ratio of gas moles

V₁/n₁=V₂/n₂

1 = reactants, 2 = products

Input the value :

5.85/0.16=V₂/0.15

V₂=5.484 L

6 0

2 years ago