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3 years ago

Why is Carbon Tetrahydride a covalent bond? plzzz helpppp!!!!!!

Physics

1 answer:

arlik [135]3 years ago

7 0

Answer:

carbon has four unpaired electrons in its valence shell . hydrogen having one unpaired electron in its valence shell comes to bond with carbon by sharing a pair of electrons .since carbon needs 4 electrons to be stable, 4 hydrogen atoms take part in the bond . It's a covalent bond because the difference between the electronegativity of carbon and hydrogen is quite small .

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Why is figure 5 an unhelpful visualization tool for this data set? Please help!

Paraphin [41]

Explanation:

Because the temperature and the radiation are not correlated, they're not represented as functions of each other, they're represented as independent variables thus using graph 5 you cannot figure out how one affect another

8 0

2 years ago

In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu

Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?

Explanation:

The given data is as follows.

Mass of proton = $1.67 \times 10^{-27}$ kg

Charge of proton = $1.6 \times 10^{-19} C$

Speed of proton = $2.50 \times 10^{6} m/s$

Distance traveled = $5.31 \times 10^{-13} m$

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

$(K.E + P.E)_{initial}$ = $(K.E + P.E)_{final}$

$(\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)$

where, $\frac{kq_{p}q_{t}}{r} = U = Electric potential energy$

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

Putting the given values into the above formula as follows.

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

= $(\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})$

= $5.218 \times 10^{-15} J$

Therefore, we can conclude that the electric potential energy of the proton and nucleus is $5.218 \times 10^{-15} J$ .

4 0

3 years ago

Tina has been dieting for a total of 13 weeks she lost 3 lb on the first week of her diet but gain the back of pound on the seco

PolarNik [594]

Answer:

D. 24 lb

Explanation:

Tina has been dieting for 13 weeks

First week she lost 3 pounds

Next week she gained 1 pound and did not lose any. This will be subtracted as she has gained a pound

The remaining 11 weeks she lost 2 pounds per week

Weight lost in the 11 weeks = 11×2 =22 pounds

Total weight lost

3-1+22 = 24 lb

Tina has lost 24 pounds in total during the 13 weeks

6 0

3 years ago

Match each letter to the description

Makovka662 [10]

A woman walks in a straight line with the sun to her right at six o'clock in the morning.

The sun rises East of her, so the woman is walking toward the North pole.

A man walks in a straight line with the sun to his right at six o'clock in the evening.

The sun sets West of him, so the man is walking toward the South pole.

The woman and the man are both walking along lines of constant longitude.

5 0

3 years ago

You are investigating how objects move when they are dropped from different heights. To collect your data, you drop a 1 kg weigh

ASHA 777 [7]

The time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

The given parameters;

Mass of the first object, m1 = 1 kg

Mass of the second object, m2 = 5 kg

The final velocity of the objects during the downward motion is calculated as follows;

$v_f = v_0 + gt\\\\v_f = 0 + gt\\\\\v_f = gt$

The time of motion of the object from the given height is calculated as;

$h = v_0 t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} }$

The time of motion of each object is independent of mass of the object.

Thus, the time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

Learn more about time of motion here: brainly.com/question/2364404

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2 years ago

Why is Carbon Tetrahydride a covalent bond? plzzz helpppp!!!!!!​

Why is Carbon Tetrahydride a covalent bond? plzzz helpppp!!!!!!