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3 years ago

10

Calculate the specific heat of a substance when 63j of energy are transferred as heat to an 8.0 g sample to raise it temperature

from 314 k to 340 k

2 answers:

lubasha [3.4K]3 years ago

8 0

Answer : The specific heat of substance is 0.30 J/g.K

Explanation :

Formula used :

$Q=m\times c\times \Delta T$

or,

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat = 63 J

m = mass of substance = 8.0 g

$C_w$ = specific heat of substance = ?

$T_1$ = initial temperature = 314 K

$T_2$ = final temperature = 340 K

Now put all the given value in the above formula, we get:

$63J=8.0g\times c\times (340-314)K$

$c=0.30J/g.K$

Therefore, the specific heat of substance is 0.30 J/g.K

Flura [38]3 years ago

6 0

The formula for energy or enthalpy is:

E = m Cp (T2 – T1)

where E is energy = 63 J, m is mass = 8 g, Cp is the specific heat, T is temperature

63 J = 8 g * Cp * (340 K – 314 K)

<span>Cp = 0.3 J / g K</span>

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Why can’t there be a temperature lower than absolute zero

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2 years ago

Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f

N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+2B\rightleftharpoons C$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first trial:

$5.4\times 10^{-3}=k(0.30)^a(0.050)^b$ ....(1)

Expression for rate law for second trial:

$1.1\times 10^{-2}=k(0.30)^a(0.100)^b$ ....(2)

Expression for rate law for third trial:

$2.2\times 10^{-2}=k(0.50)^a(0.050)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1$

Dividing 3 by 1, we get:

$\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^1$ ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}$

Calculating the initial rate of formation of C by using equation 4, we get:

$k=1.2M^{-2}s^{-1}$

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

$\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}$

Hence, the initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

8 0

3 years ago

Suppose that a certain biologically important reaction is quite slow at physiological temperature (37 oC) in the absence of a ca

Oksi-84 [34.3K]

Answer:

30 kJ

Explanation:

Arrhenius equation is given by:

$k=Aexp(-Ea/RT)\\$

Here, k is rate constant, A is Pre-exponential factor, Ea is activation energy and T is temperature.

taking natural log of both side

ln k = ln A - Ea/RT

In Arrhenius equation, A, R and T are constant.

Therefore,

$ln\frac{k_2}{k_1} =\frac{Ea_1-Ea_2}{RT}$

$Ea_1-Ea_2$ is the lowering in activation energy by enzyme,

R = 8.314 J/mol.K

T = 37°C + 273.15 = 310 K

$\frac{k_2}{k_1} =1\times 10^5$

$ln 1\times 10^5 =\frac{Ea_1-Ea_2}{RT}\\{Ea_1-Ea_2} = 11.512 \times 8.314 \times 310\\=29670\ J\\=30\ kJ$

4 0

3 years ago