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3 years ago

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How many bonds can a carbon atom form

2 answers:

White raven [17]3 years ago

6 0

A carbon atom can form the following bonds: Four single bonds.

Eight valence electrons need to be present in carbon's outer shell when considering covalent bonding. The three major types of covalent bonds are single, double, and triple bonds.

kotegsom [21]3 years ago

3 0

A carbon atom can form four bonds.

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F = 2.0*10^20 N,

natka813 [3]

$F=\dfrac{Gm_1m_2}{r^2}$

With the given values of $F,G,m_1,r$ , we have

$2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}$

Try dealing with the powers of 10 first: On the right, we have

$\dfrac{10^{-11}\times10^{24}}{(10^8)^2}=\dfrac{10^{24-11}}{10^{16}}=10^{-3}$

Meanwhile, the other values on the right reduce to

$\dfrac{6.67\times5.98}{3.8^2}\approx2.76$

Then taking units into account, we end up with the equation

$2.0\times10^{20}\,\mathrm N=\left(2.76\times10^{-3}\,\dfrac{\mathrm N}{\mathrm{kg}}\right)m_2$

Now we solve for $m_2$ :

$m_2=\dfrac{2.0\times10^{20}\,\mathrm N}{2.76\times10^{-3}\,\frac{\mathrm N}{\mathrm{kg}}}\approx0.725\times10^{20-(-3)}\,\mathrm{kg}$

$m_2=7.25\times10^{22}\,\mathrm{kg}$

or, if taking significant digits into account,

$m_2=7.3\times10^{22}\,\mathrm{kg}$

5 0

3 years ago

Below is a single strand of DNA, what is the complementary base pair for the strand?

GREYUIT [131]

The complementary base pair is:
TTC, CTG, AGT, CTA.

5 0

3 years ago

The temperature of a body is from 200 to 300C.The change of temp at absolute scale is

Oliga [24]

Answer:

mark me brainliest

Explanation:

The change of temperature at absolute scale is. A. 3.73 K

7 0

3 years ago

Read 2 more answers

In a given chemical reaction, the energy of the products is kess than the energy of the reactants. Which statement is true for t

deff fn [24]

Answer: Released

Explanation: Energy is released in this reaction possibly in the form of heat thus it is an exergonic and or exothermic reaction.

5 0

3 years ago

There is a plate with moment of inertia Ip = 0.0711 kgm2 , rotating around its center of mass with angular speed of 4.53 rad/s.

Anna007 [38]

Answer:

1) their common angular speed = 3.038 rad/s

2) kinetic energy loss = 0.24J

Explanation:

1) This is a case of conservation of angular momentum.

The initial angular momentum must be equal to the final angular momentum

Initial angular momentum = I x w

Where I = moment of inertia, and

w = angular momentum.

Initial angular momentum = 0.0711 x 4.53 = 0.322 kg-m2-rad/s

After addition of ring, moment of inertia becomes,

0.0711 + 0.0353 = 0.106 kg-m2

Therefore, final angular momentum will be

0.106 x Wf

Where Wf = final common angular velocity

Equating the two angular momentum we have

0.322 = 0.106Wf

Wf = 0.322/0.106 = 3.038 rad/s

2) KE = 1/2 x I x w^2

Initial KE = 1/2 x 0.0711 x 4.53^2

= 0.729 J

Final KE = 1/2 x 0.106 x 3.038^2

= 0.489 J

Loss in KE = 0.729 - 0.489 = 0.24 J

5 0

4 years ago