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2 years ago

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A 8.00g sample of substance (substance, molar mass = 152.0 g/mol) was combusted in a bomb calorimeter with a heat capacity of 6.

21 kJ/°C. If the temperature rose from 25.0°C to 75.0°C, determine the value of ΔH°comb for substance.

1 answer:

aleksandrvk [35]2 years ago

4 0

Answer:

ΔH°comb=-5899.5 kJ/mol

Explanation:

First, consider the energy balance:

$m_{c} *Cp*(T_{2}-T_{1})=-n_{s} *H_{c}$ Where $m_{c}$ is the calorimeter mass and $n_{s}$ is the number of moles of the samples; $H_{c}$ is the combustion enthalpy. The energy balance says that the energy that the reaction release is employed in rise the temperature of the calorimeter, which is designed to be adiabatic, so it is suppose that the total energy is employed rising the calorimeter temperature.

The product $m_{c} *Cp$ is the heat capacity, so the balance equation is:

$6.21\frac{kJ}{K}*(75-25)=-8.00g*\frac{mol}{152.0g}*H_{c}$

So, the enthalpy of combustion can be calculated:

$H_{c}=-5899.5\frac{kJ}{mol}$

I will be happy to solve any doubt you have.

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2 years ago

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Evaluate the final kinetic energy of the supply spacecraft for the actual tractor beam force, F(x)=αx3+βF(x)=αx3+β.

Elan Coil [88]

Answer:

K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x

Explanation:

To find the variation of kinetic energy, let's use the work energy theorem

W = ΔK

∫ F .dx = K -K₀

If the body starts from rest K₀ = 0

∫ F dx cos θ = K

Since the force and displacement are in the same direction, the angle is zero, so the cosine is 1

we substitute and integrate

α ∫ x³ dx + β ∫ dx = K

α x⁴ / 4 + β x / 1 = K

we evaluate from the lower limit F = 0 to the upper limit F

α (x⁴ / 4 -0) + β (x -0) = K

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2 years ago

A particle with a charge of +4.20nC is in a uniform electric field E⃗ directed to the left. It is released from rest and moves t

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Answer:

(A). The work done is $1.50\times10^{-6}\ J$ .

(B). The potential of the starting point with respect to the endpoint is 357.14 V.

(C). The magnitude of E is 5952.38 N/C.

Explanation:

Given that,

Charge = 4.20 nC

Distance = 6.00 cm

Kinetic energy $K.E=1.50\times10^{-6}\ J$

The particle start from rest.

So, the initial kinetic energy i zero.

(A). We need to calculate the work by the electric force

Using formula of work done

$W = \Delta K.E$

$W=K.E_{f}-K.E_{i}$

Put the value into the formula

$W= 1.50\times10^{-6}-0$

$W=1.50\times10^{-6}\ J$

The work done is $1.50\times10^{-6}\ J$ .

(B). We need to calculate the potential of the starting point with respect to the endpoint

We know that.

Change in potential energy = change in kinetic energy

$\Delta P.E=\Delta K.E$

So, $U = 1.50\times10^{-6}$

Using formula of potential

$V=\dfrac{U}{q}$

Put the value into the formula

$V=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}}$

$V=357.14\ V$

The potential of the starting point with respect to the endpoint is 357.14 V.

(C). We need to calculate the magnitude of E

Using formula of work done

$W=F\times r$ ....(I)

Using formula of force

$F=qE$

Put the value in the equation (I)

$W=qE\times r$

$E=\dfrac{W}{q\times r}$

Put the value into the formula

$E=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}\times6.00\times10^{-2}}$

$E=5952.38\ N/C$

The magnitude of E is 5952.38 N/C.

Hence, This is the required solution.

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2 years ago

A fisherman casts his bait toward the river at an angle of 25° above the horizontal. As the line unravels, he notices that the b

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Answer:

$v_{o}=18m/s$

Explanation:

According to the exercise we know the angle which the bait was released and its maximum height

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$v_{y}^{2} =v_{oy}^{2}+2g(y-y_{o})$

At maximum height the y-component of velocity is 0 and from the exercise we know that the initial y position is 0

$0=v_{oy}^{2}-2(9.8m/s^2)(2.9m)$

$v_{oy}=\sqrt{2(9.8m/s^{2} )(2.9m)} =7.53m/s$

Since the bait is released at 25º

$v_{oy}=v_{o}sin(25)$

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1 year ago

Two point charges totaling 8 μC exert a repulsive force of 0.15 N on one another when separated by 0.5 m. What is the charge on

nexus9112 [7]

Answer:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

Explanation:

From Coulomb's Law the electrostatic repulsive force is given by the following formula:

F = kq₁q₂/r²

where,

F = Repulsive Force = 0.15 N

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = Magnitude of 1st Charge = ?

q₂ = Magnitude of 2nd Charge = ?

r = Distance between Charges = 0.5 m

Therefore,

0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²

q₁q₂ = (0.15 N)(0.5 m)²/(9 x 10⁹ N.m²/C²)

q₁q₂ = 4.17 x 10⁻¹²

q₁ = (4.17 x 10⁻¹²)/q₂ -------------------- equation (1)

The sum of charges is given as:

q₁ + q₂ = 8 μC

q₁ + q₂ = 8 x 10⁻⁶

using equation (1):

(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶

(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂

q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0

Solving this quadratic equation:

q₂ = 7.4 x 10⁻⁶ C (OR) q₂ = 0.56 x 10⁻⁶ C

q₂ = 7.4 μC (OR) q₂ = 0.6 μC

Therefore,

q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)

q₁ = 0.6μC

Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.

Therefore, the correct option will be:

<u>B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C</u>

8 0

2 years ago