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2 years ago

How many moles of MgCl2 are there in 302 g of the compound?

Physics

2 answers:

VashaNatasha [74]2 years ago

8 0

MgCl2 molar mass is about 94g/mol so 302/94 is about 3.2. So the answer is 3.2 moles in 302 g.

Hope this helps:)

lisov135 [29]2 years ago

3 0

Answer:

‍♀️‍♀️‍♀️‍♀️sorrryyyyy

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The coil of an ac generator has 50 loops and a cross-sectional area of LaTeX: 0.2~m^20.2 m 2 . What is the maximum emf that can

lisov135 [29]

Answer:

Maximum emf in the coil is 100 volt

Explanation:

We have given number of loops in the coil N = 50

Cross sectional area $A=0.2m^2$

Angular speed is given $\omega =5rad/sec$

Magnetic field is given $B=0.2T$

We have to find the maximum emf in the coil

Maximum emf in the coil is given by $e=NBA\omega$

$e=50\times 2\times 0.2\times 5=100volt$

So maximum emf induced in the coil is 100 volt

7 0

3 years ago

A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the a

DENIUS [597]

Answer:

C = 4,174 10³ V / m^{3/4} , E = 7.19 10² / ∛x, E = 1.5 10³ N/C

Explanation:

For this exercise we can calculate the value of the constant and the electric field produced,

Let's start by calculating the value of the constant C

V = C $x^{4/3}$

C = V / x^{4/3}

C = 220 / (11 10⁻²)^{4/3}

C = 4,174 10³ V / m^{3/4}

To calculate the electric field we use the expression

V = E dx

E = dx / V

E = ∫ dx / C x^{4/3}

E = 1 / C x^{-1/3} / (- 1/3)

E = 1 / C (-3 / x^{1/3})

We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E

E = 3 / C (0- (-1 / x^{1/3}))

E = 3 / 4,174 10³ (1 / x^{1/3})

E = 7.19 10² / ∛x

for x = 0.110 cm

E = 7.19 10² /∛0.11

E = 1.5 10³ N/C

6 0

3 years ago

Given a 10-V power supply, would a 20-ohm resistor and a 5-ohm resistor need to be arranged in parallel or in series to generate

Bogdan [553]

Answer:

The resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.

Explanation:

We are given 10 voltage power source and we have two Resistors with resistance of 20 ohm and 5ohm.

We need to find the orientation in which these two resistors would be arranged so that the circuit could get a current of 2.5Ampere.

Using ohm's law we have

V = I*R

V= voltage

I= current

R= resistance

10 = 2.5*R

R = 10/2.5 = 4ohm

that means we need a total of 4ohm resistance from these two resistors.

since the net Resistance(4ohm) is lower than the smallest resistance(5ohm) available that means the orientation of the resistors will be in parallel.

$\frac{1}{R} = \frac{1}{R1} +\frac{1}{R2}\\ = \frac{1}{20} +\frac{1}{5}\\ \\ = \frac{1+4}{20} =\frac{1}{4}$

R(net) =4ohm

Now the orientation of the resistors are in parallel so the current will be divided.

we know that the current will divide in opposite manner the arm which provides more resistance less current will flow from there and vice versa.

We know that the voltage in parallel remains same

In 20 ohm resistance

again using ohms law

V = i1*R1

10 = i1*20

i1 = 0.5A

in 5ohm resistor

V=i2*R2

10 = I2*5

i2 =2A

and i1+i2 = 0.5+2= 2.5A which means our calculation is correct.

Therefore the resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.

6 0

2 years ago

A 60.00-cm guitar string under a tension of 50.000 N has a mass per unit length of 0.100 00 g/cm chegg

Evgen [1.6K]

Fundamental frequency,

f=v2l=T/μ−−−−√2l

=(50)/0.1×10−3/10−22×0.6−−−−−−−−−−−−−−−−−−−√

=58.96Hz

Let, n th harmonic is the hightest frequency, then

(58.93)n = 20000

∴N=339.38

Hence, 339 is the highest frequency.

∴fmax=(339)(58.93)Hz=19977Hz.

<h3>What is frequency?</h3>

In physics, frequency is the number of waves that pass a given point in a unit of time as well as the number of cycles or vibrations that a body in periodic motion experiences in a unit of time. After moving through a sequence of situations or locations and then returning to its initial position, a body in periodic motion is said to have experienced one cycle or one vibration. See also simple harmonic motion and angular velocity.

learn more about frequency refer:

brainly.com/question/254161

#SPJ4

7 0

1 year ago

A mass m attached to a horizontal massless spring with spring constant k, is set into simple harmonic motion. its maximum displa

Lesechka [4]

At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:
$U_i= \frac{1}{2} ka^2$
while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:
$K_i =0$
And the total energy of the system is
$E_i = U_i+K= \frac{1}{2}ka^2$

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:
$U_f = 0$
while the mass is moving at speed v, and therefore the kinetic energy is
$K_f = \frac{1}{2} mv^2$
And the total energy is
$E_f = U_f + K_f = \frac{1}{2} mv^2$

For the law of conservation of energy, the total energy must be conserved, therefore $E_i = E_f$ . So we can write
$\frac{1}{2} ka^2 = \frac{1}{2}mv^2$
that we can solve to find an expression for v:
$v= \sqrt{ \frac{ka^2}{m} }$

6 0

3 years ago