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3 years ago

13

Assume a reaction takes 7.5 moles of anhydrous calcium chloride and energy transfer occurs, which we record as 21.2 joules of en

ergy. Find the heat of solution in J/mol and round to two places after the decimal point.

2 answers:

OLga [1]3 years ago

8 0

Answer: 2.83 J/mol

Explanation:

Heat of solution, sometimes interchangeably called enthalpy of solution, is said to be the energy released or absorbed when the solute dissolves in the solvent. A solute is that which can dissolve in a solvent, to form a solution

Given

No of moles of CaCl = 7.5 mol

Total energy used = 21.2 J

Heat of solution = q/n where

q = total energy

n = number of moles

Heat of solution = 21.2 / 7.5

Heat of solution = 2.83 J/mol

netineya [11]3 years ago

3 0

Answer:

Heat of solution; ΔH = 2.83 J/mol

Explanation:

We are given;

Number of moles; n = 7.5 moles

Energy absorbed; q = 21.2 J

Now, in endothermic reactions,

ΔH = q/n

Where ;

ΔH is heat of solution

q is amount of energy absorbed

n is number of moles of solute

Thus, plugging in relevant values, we have;

ΔH = q/n

ΔH = 21.2/7.5

ΔH = 2.83 J/mol

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Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?

Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = $\frac{1}{f}$

where

f = focal length

Thus

f = $\frac{1}{P}$

f = $\frac{1}{2}$ = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}$

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

$\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}$

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0

3 years ago

is this true or false? an object with a mass of 9.0kg has an acceleration of 3m/s2. the force acting on it is 3N

ludmilkaskok [199]

The answer is "False". The force acting on the object is 27 N.

According to Newton's second law, when a force <em>F</em> acts on am object of mass <em>m</em>, it produces an acceleration <em>a</em>. The force is given by the expression,

$F=ma$

Thus, if the body has a mass of 9.0 kg and if it has an acceleration of 3 m/s², then, on substituting the values in the equation for force,

$F=ma\\ =(9.0kg)(3m/s^2)\\ =27N$

Thus, it can be seen that the force acting on the body is 27 N and not 3 N as is mentioned in the statement. Hence the statement is false.

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3 years ago

In January 2006, astronomers reported the discovery of a planet comparable in size to the earth orbiting another star and having

Orlov [11]

Answer:

R = 5.28 103 km

Explanation:

The definition of density is

ρ = m / V

V = m /ρ

Where m is the mass and V the volume of the body

The volume of a sphere is

V = 4/3 π r³

Let's replace

4/3 π r³ = m / ρ

R =∛ ¾ m / ρ π

The mass of the planet is

M = 5.5 Me

R = ∛ ¾ 5.5 Me /ρ π

Let's reduce the density to SI units

ρ = 1.76 g / cm³ (1 kg / 10³ g) (10² cm / 1 m)³

ρ = 1.76 10³ kg / m³

Let's calculate

R = ∛ ¾ 5.5 5.97 10²⁴ / (1.76 10³ pi)

R = ∛ 0.14723 10²¹

R = 0.528 10⁷ m

R = 0.528 104 km

R = 5.28 103 km

8 0

3 years ago

How does an electromagnet become permanent

love history [14]

<span>A moving electrical charge produces a magnetic field and a moving magnetic field produces an electrical field. An electromagnet works by coiling a bunch of wire and spinning a couple of magnets around that wire at high speeds. When this occurs the magnets induce an electric current in the wire and hence the electricity production. Once the magnets stop spinning, the induced electrical field dissipates and the current stops flowing through the wire.

</span>

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3 years ago

A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$

Differentiating with respect to t ( time ),

$0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ ( l = 26 feet = constant )

$\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}$

$\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

We have,

$y = 10, \frac{dx}{dt}= -3\text{ feet per min}$

$\frac{dy}{dt}=\frac{3x}{10}-----(X)$

(i) From equation (1),

$26^2 = x^2 + 10^2$

$676=x^2 + 100$

$576 = x^2$

$\implies x = 24\text{ feet}$

From equation (X),

$\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}$

(ii) From equation (X),

$\frac{dy}{dt}\propto x$

Thus, for different value of x the value of $\frac{dy}{dt}$ would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

$\frac{dy}{dt}=0\text{ feet per min}$

3 0

3 years ago