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kykrilka [37]
3 years ago
7

Write on the discovery of electricity not less than a page​

Physics
1 answer:
Veronika [31]3 years ago
8 0

Cevabı bilmiyom arkadaşım benn türküm puan almak için cevap veriyorum hadi bbyy

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If six moles of hydrogen chloride (HCl) react with plenty of aluminum, how many moles of aluminum chloride (AlCl3) will the reac
AlexFokin [52]

Answer:

Two moles of aluminum chloride (AlCl_3) are produced when six miles of hydrogen Chloride (HCl) react with plenty of aluminum

Explanation:

6 Moles of HCl will only react with 2 moles of Al irrespective of the number of moles of each compound present. The reaction wiil take place in this ratio only. The products produced will be 2 moles of AlCl_3 and 3 moles of H_2 this ratio will also be constant.

So, six moles of hydrogen chloride (HCl) will react with plenty of aluminum to produce many 2 moles of aluminum chloride (AlCl_3).

5 0
3 years ago
Which molecule will undergo only london dispersion forces when interacting with other molecules of the same kind?
vfiekz [6]

Explanation:

London dispersion forces will form between non-polar molecules(polar ) that are symmetrical  like O₂, H₂, Cl₂ and noble gases.

  • The attraction here is because non-polar molecules becomes polar due to the constant motion of its electrons.
  • This lead to an uneven charge distribution at an instant.
  • A temporary  dipole or instantaneous dipole forms.
  • The temporary dipole can induce neighboring molecules to be distorted and forms dipoles as well.
  • This forms london dispersion forces.

Learn more:

Intermolecular forces brainly.com/question/10602513

#learnwithBrainly

7 0
3 years ago
Tutorial Exercise An unstable atomic nucleus of mass 1.83 10-26 kg initially at rest disintegrates into three particles. One of
kogti [31]

Answer:

A) v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) K_total = 373.08 × 10^(-15) J

Explanation:

We are given;

Mass of unstable atomic nucleus; M = 1.83 × 10^(-26) kg

Mass of first particle; m1 = 5.03 × 10^(-27) kg

Speed of first particle in y-direction; v1 = (6 × 10^(6) m/s) j^

Mass of second particle; m2 = 8.47 × 10^(-27) kg

Speed of second particle in x - direction; v2 = (4 × 10^(6) m/s) i^

Now, we don't have the mass of the third particle but since we are told the unstable atomic nucleus disintegrates into 3 particles, thus;

M = m1 + m2 + m3

1.83 × 10^(-26) = (5.03 × 10^(-27)) + (8.47 × 10^(-27)) + m3

m3 = (1.83 × 10^(-26)) - (13.5 × 10^(-27))

m3 = 4.8 × 10^(-27) kg

A) Applying law of conservation of momentum, we have;

MV = (m1 × v1) + (m2 × v2) + (m3 × v3)

Now, the unstable atomic nucleus was at rest before disintegration, thus V = 0 m/s.

Thus, we now have;

0 = (m1 × v1) + (m2 × v2) + (m3 × v3)

We want to find the velocity of the third particle v3. Let's make it the subject of the formula;

v3 = [(m1 × v1) + (m2 × v2)]/(-m3)

Plugging in the relevant values, we have;

v3 = [(5.03 × 10^(-27) × 6 × 10^(6))j^ + (8.47 × 10^(-27) × 4 × 10^(6))i^]/(-4.8 × 10^(-27))

v3 = [(30.18 × 10^(-21))j^ + (33.88 × 10^(-21))i^]/(-4.8 × 10^(-27))

v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) Formula for kinetic energy is;

K = ½mv²

Now,total kinetic energy is;

K_total = K1 + K2 + K3

K1 = ½ × 5.03 × 10^(-27) × (6 × 10^(6))²

K1 = 90.54 × 10^(-15) J

K2 = ½ × 8.47 × 10^(-27) × (4 × 10^(6))²

K2 = 67.76 × 10^(-15)

To find K3, let's first find the magnitude of v3 because it's still in vector form.

Thus;

v3 = √[(-6.29 × 10^(6))² + (-7.06 × 10^(6))²]

v3 = 9.46 × 10^(6) m/s

K3 = ½ × 4.8 × 10^(-27) × (9.46 × 10^(6))²

K3 = 214.78 × 10^(-15) J

K_total = (90.54 × 10^(-15)) + (67.76 × 10^(-15)) + (214.78 × 10^(-15))

K_total = 373.08 × 10^(-15) J

7 0
3 years ago
As you learned in Part B, a non-burning helium core surrounded by a shell of hydrogen-burning gas characterizes the subgiant sta
Gennadij [26K]

Answer:

E- The star becomes a red giant (LATEST STAGE)

F- The surface of the star becomes brighter and cooler

C- Pressure from the star's hydrogen-burning shell causes the non burning envelope to expand

A- The shell of hydrogen surrounding the star's nonburning helium core ignites.

D- The star's non burning helium core starts to contract and heat up

B- Pressure in the star's core decreases (EARLIEST STAGE)

(A star moves away from the main sequence once its core runs out of hydrogen to fuse into helium. The energy once supplied by hydrogen burning reduces and the core starts to compress under the force of gravity. This contraction allows the core and surrounding layers to heat up. Finally, the hydrogen shell around the core becomes hot enough to ignite hydrogen burning.

6 0
3 years ago
Una patinadora de 50 kg parte del reposo y después de recorrer 3k alcanza una velocidad de 15 m/s. ¿Qué fuerza neta experimenta
e-lub [12.9K]

Answer:

F_{net} = 1.875\,N

Explanation:

Asúmase que la patinadora experimenta una aceleración constante. La fuerza neta experimentada por la patinadora:

F_{net} = (50\,kg)\cdot \left[\frac{\left(15\,\frac{m}{s}\right)^{2}-\left(0\,\frac{m}{s}\right)^{2} }{2\cdot (3000\,m)} \right]

F_{net} = 1.875\,N

6 0
3 years ago
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