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3 years ago

Explain why protons attached to nitrogen (i.e. n-h show broad peaks in the 1h nmr and do not couple to other protons.

1 answer:

7 0

The element Nitrogen is usually present in almost all proteins, and it also forms strong bonds due to its ability to form a triple bond with its self, also to other elements.

<span>The nitrides use a variety of different oxidation and almost all the oxides forms gasses are at 25 degrees Celsius. The oxide of nitrogen are acidic, thus, easily attaching protons.</span>

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On august 10, 1972, a large meteorite skipped across the atmosphere above the western united states and western canada, much lik

RUDIKE [14]

(a)
The velocity of the meteorite just before hitting the ground is:
$v=20 km/s=20000 m/s$
The loss of energy of the meteorite corresponds to the kinetic energy the meteorite had just before hitting the ground, so:
$\Delta K = \frac{1}{2}mv^2= \frac{1}{2}(3.4 \cdot 10^6 kg)(20000 m/s)^2=6.8 \cdot 10^{14}J$

(b) 1 megaton of tnt is equal to $1 MT=4.2 \cdot 10^{15}J$
To find to how many megatons the meteorite energy loss $\Delta E$
corresponds, we can set the following proportion
$1 MT: 4.2 \cdot 10^{15}J=x: \Delta E$
And so we find
$x= \frac{\Delta E}{4.2 \cdot 10^{15}J} = \frac{6.8 \cdot 10^{14}J }{4.2 \cdot 10^{15}J} =0.162 MT$
So, 0.162 megatons.

(c) 1 Hiroshima bomb is equivalent to 13 kilotons (13 kT). The impact of the meteorite had an energy of $\Delta E=0.162 MT=162 kT$ . So, to find to how many hiroshima bombs it corresponds, we can set the following proportion:
$1:13 kT=x:162 kT$
And so we find
$x= \frac{162 kT}{13 kT}=12.46$
So, the energy released by the impact of the meteorite corresponds to the energy of 12.46 hiroshima bombs.

7 0

4 years ago

The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just bare

Brums [2.3K]

Answer:

6862.96871 seconds

Explanation:

M = Mass of Planet

G = Gravitational constant

r = Radius

$\rho$ = Density

T = Rotation period

In this system the gravitational force will balance the centripetal force

$G\frac{Mm}{r^2}=mr\omega^2$

$\omega=\frac{2\pi}{T}$ .

$M=\rho v\\\Rightarrow M=\rho \frac{4}{3}\pi r^3$

$\\\Rightarrow G\frac{Mm}{r^2}=mr\left(\frac{2\pi}{T}\right)^2\\\Rightarrow \frac{G\rho \frac{4}{3}\pi r^3}{r^3}=\frac{4\pi^2}{T^2}\\\Rightarrow T=\sqrt{\frac{3\pi}{G\rho}}$

Hence, proved

$T=\sqrt{\frac{3\pi}{6.67\times 10^{-11}\times 3000}}\\\Rightarrow T=6862.96871\ s$

The rotation period of the astronomical object is 6862.96871 seconds

7 0

3 years ago

Solids in which the atoms have no particular order or pattern are called

posledela

Answer:

Amorphous solids are composed of atoms or molecules that are in no particular order. Each particle is in a particular spot, but the particles are in no organized pattern. Examples include rubber and wax. Crystalline solids have a very orderly, three-dimensional arrangement of atoms or molecules

Explanation:

8 0

3 years ago

The flow of electric current through a gas without any external influence (ionizer) is called

maksim [4K]

Answer:

self-maintained discharge

Explanation:

5 0

3 years ago

determine the pressure exerted on the surface of a submarine cruising 175 ft below the free surface of the sea. assume that the

Alenkasestr [34]

Answer:

92.81 psia.

Explanation:

The density of water by multiplying its specific gravity by the density of sea water.

SG = density of sea water/density of water

ρ = SG x ρw

1 kg/m3 = 62.4 lbm/ft^3

= 1.03 * 62.4

= 64.27lbm/ft^3.

The absolute pressure at 175 ft below sea level as this is the location of the submarine.

P = Patm +ρgh

= 14.7 + 64.27 * 32.2 * 175

Converting to pound force square inch,

= 14.7 + 64.27 * (32.2ft/s^2) * (175ft) * (1lbf/32.2lbm⋅ft/s^2) * (1ft^2/144in^2 )

= 14.7 + 78.11 psia

= 92.81 psia.

8 0

3 years ago