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Ann [662]
2 years ago
10

Hydrogen bonds are too weak to bind atoms together to form molecules, but they do hold different parts of a single large molecul

e in a specific three-dimensional shape. true false
Physics
1 answer:
pogonyaev2 years ago
5 0

Hydrogen bonds are too weak to bind atoms together to form molecules, but they do hold different parts of a single large molecule in a specific three-dimensional shape. The given statement is true.

<h3>What are hydrogen bonds?</h3>

A hydrogen bond is an electrostatic force of attraction among a hydrogen atom tightly attached to a more electronegative "donor" atom or group and another electronegative atom bearing a lone pair of electrons, known as the hydrogen bond acceptor.

Hydrogen bonds are too flimsy to connect atoms to form molecules, but they do hold various portions of a single large molecule together in a specific three-dimensional shape.

Thus, the given statement is true.

For more details regarding hydrogen bonding, visit:

brainly.com/question/10904296

#SPJ1

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What is an “equilibrium temperature”?
RSB [31]

Answer:

One may assume that the planet radiates energy like a blackbody at some temperature according to the Stefan–Boltzmann law. Thermal equilibrium exists when the power supplied by the star is equal to the power emitted by the planet. The temperature at which this balance occurs is the planetary equilibrium temperature.

Explanation:

6 0
3 years ago
Carlos is making phosphorous trichloride using the equation below. He adds 15 g of phosphorus.
Degger [83]
Given:

The balanced chemical reaction of the synthesis of phosphorus trichloride:

2P + 3Cl2 ===> 2PCl3

Initial amount of phosphorus = 15 grams

The amount of product produced from 15 grams of phosphorus:

15 grams / 31 g/mol * (2/2) = 66.46 grams PCl3 

The amount of chlorine is 44.31 grams, nearest to 45 grams. 
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4 years ago
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A (blank) is a point on a standing wave that appears to be stationary
Vikentia [17]

Hi! Your answer would be "node"

7 0
3 years ago
Tidal Forces near a Black Hole. An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black
arlik [135]

Answer:

833.4801043*10^6N on ear that is closer to Black hole.

13.83803929*10^6N On ear that is farther from Black hole.

Explanation:

This problem can be solved as two masses that are at two different location from a bigger mass whose gravity affects both.

tension is an equal and opposite force that is exerted in response to applied force.

so on ear that is closer to black hole would have tension that is equal in magnitude and opposite in direction to gravitational force that ear experience due to the black hole at that location.

this true for ear that is further away from black hole as well.

(1) Force on ear that is closer to black hole.

                                                 F =\frac{m_{1}*m_{2}  }{r^2} G

                    m_{1}= 5*1.989*10^30kg is mass of Black hole.

                     m_{2}= 0.030kg is mass of ear that is close to black hole.

                    G = 6.679*10^-11m^3*kg^-1*s^-2

                     r = 120km-\frac{6}{100000} km=119.99994km=119999.94m

Note, we have subtracted because ear is closer to black hole.

plugging all this in formula gives.

                                      F = 833.4801043*10^6N

       That is tension of ear.

(2) Force on ear that is further from black hole.

                              F =\frac{m_{1}*m_{2}  }{r^2} G

                    m_{1}= 5*1.989*10^30kg is mass of Black hole.

                     m_{2}= 0.030kg is mass of ear that is close to black hole.

                    G = 6.679*10^-11m^3*kg^-1*s^-2

         this time r is further away from black hole so it would be.

                       r = 120km+\frac{6}{100000} km = 120000.06m

Plugging this all in we get

                          F = 13.83803929N

and that is tension on ear that is further from black hole.

Notice the tension difference, and order of magnitude of tension,it is enormous .

this astronaut is lethally close to black hole.

5 0
3 years ago
Gauss's law combines the electric field over a surface with the area of the surface. From Coulomb's law we know that the electri
Romashka-Z-Leto [24]

The change in surface area of Gaussian surface with radius (r) is 8πr.

<h3>Electric field from Coulomb's law</h3>

The electric field experienced by a charge is calculated as follows;

E = \frac{Q}{4\pi \varepsilon_o r^2}

where;

  • E is the electric field
  • Q is the charge
  • r is the radius

The electric field reduces by a factor of \frac{1}{r^2}

<h3>Surface area of a Gaussian surface;</h3>

The surface area of a sphere is given as;

A = 4\pi r^2

<h3>Change in area with r</h3>

\frac{dA}{dr} = 8\pi r

Thus, the change in surface area of Gaussian surface with radius (r) is 8πr.

Learn more about area of Gaussian surfaces here: brainly.com/question/17060446

7 0
3 years ago
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