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PtichkaEL [24]
2 years ago
7

What is the chemical formula for copper (I) sulfide? A) CuS B) CuS2 C) Cu2S D) Cu2S3

Physics
2 answers:
Ganezh [65]2 years ago
4 0

This problem can be solved by using oxidation states. The oxidation state  is based on the electrons an atom loses or gain, or appears to use when forming compounds.To find the formula for Copper (I) Sulfide, we have to first consider the individual ions that come together to form it. Copper (I) is an ion of copper that has a charge of  +1, i.e  Cu^{+} . Sulfur is an element in group 6 of the periodic table. The sulfur ion has a charge of  -2, i.e  S^{2-}.  To form Copper (I) Sulfide, which is electrically neutral, we need 2 Copper (I) ions and 1 sulfur ion.

Cu_2S= 2Cu^+ 1S^-^2= 2(+1) + (-2) = 0.

The reaction is represented by the equation below

2Cu^{+} +S^{2-}-->Cu_2S.

The correct answer is (C)

grin007 [14]2 years ago
3 0

Answer:

CU2S

Explanation:

I took it just now and it is correct

You might be interested in
A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un
Kamila [148]

Answer:

<em>a) 6738.27 J</em>

<em>b) 61.908 J</em>

<em>c)  </em>\frac{4492.18}{v_{car} ^{2} }

<em></em>

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

I = \frac{1}{2}mr^{2}

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

I =  \frac{1}{2}*11*1.1^{2} = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = Iw^{2} = 6.655 x 31.82^{2} = <em>6738.27 J</em>

<em></em>

b) second flywheel  has

radius = 2.8 m

mass = 16 kg

moment of inertia is

I = \frac{1}{2}mr^{2} =  \frac{1}{2}*16*2.8^{2} = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = Iw = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

(I_{1} +I_{2} )w

where the subscripts 1 and 2 indicates the values first and second  flywheels

(I_{1} +I_{2} )w = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = Iw^{2} = 6.655 x 3.05^{2} = <em>61.908 J</em>

<em></em>

<em></em>

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = \frac{1}{2}mv_{car} ^{2}

where m is the mass of the car

v_{car} is the velocity of the car

Equating the energy

2246.09 =  \frac{1}{2}mv_{car} ^{2}

making m the subject of the formula

mass of the car m = \frac{4492.18}{v_{car} ^{2} }

3 0
3 years ago
Which of the following best explains why snow predictions by meteorologists are sometimes incorrect?
den301095 [7]

Answer:

C. Interference from the sun causes data to be collected inaccurately.

Explanation:

Snow predictions by meteorologists are sometimes incorrect because from the sun causes data to be collected inaccurately.

6 0
2 years ago
What is the average power consumption in watts of an appliance that uses 5.00 kWh of energy per day? How many joules of energy d
denpristay [2]

Answer:

(A)  power  = 0.208 kW = 208 watts

(B)  energy = 6.6 x 10^{9} joules

Explanation:

energy consumed per day = 5 kWh

(a) find the power consumed in a day

         1 day = 24 hours

        power = \frac{energy}{time}

        power = \frac{5}{24}

          power  = 0.208 kW = 208 watts

         

(b) find the energy consumed in a year

    assuming it is not a leap year and number of days = 365 days

     1 year = 365 x 24 x 60 x 60 = 31,536,000 seconds

            energy = power x time

            energy = 208 x 31,536,000

            energy = 6.6 x 10^{9} joules

5 0
3 years ago
The measure of a spring’s resistance to being compressed or stretched is the
larisa86 [58]

<u>Answer;</u>

<em>Spring constant </em>

<u>Explanation;</u>

The measure of a spring’s resistance to being compressed or stretched is the <u>spring constant</u>.

  • The symbol of spring constant is K, since it is a constant. From the Hooke's law,for a helical spring or any elastic material, the extension force is directly proportional to the extension provided the elastic limit is not exceeded.
  • Therefore; the spring constant = Force/extension. That is; K = F/e; where k is the spring constant, F is the extension force and e is the extension.
  • Spring constant depicts the resistance of the spring to compressional and stretching forces.
7 0
3 years ago
Read 2 more answers
how much force is needed to move a brick with a mass of 345-kg a distance of 28m while doing 1008 j of work?
Travka [436]

    Work = (force) x (distance)

     1,008 J  =  (force) x (28 m)

Divide each side by 28m  :    (1,008 kg-m²/sec²) / (28 m)  =  force

                                           Force =  36 kg-m/s²  =  36 Newtons .
                                                                      (about 8.1 pounds)

It doesn't matter what that force accomplishes.
It could be moving a brick, lifting a fish, or pushing a little red wagon.
In order to do 1,008 joules of work in 28 meters, it takes 36 N of force,
in the direction of the 28 meters.
5 0
2 years ago
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