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3 years ago

What is an element that needs to lose or gain 4 valence electrons in order to become stable?

Physics

1 answer:

tiny-mole [99]3 years ago

4 0

Answer:

Elements in Group 14 could lose four, or gain four electrons to achieve a noble gas structure. In fact, if they are going to form ions, Group 14 elements form positive ions. Carbon and silicon form covalent bonds. Carbon's millions of organic compounds are all based on shared electrons in covalent bonds.

Explanation:

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In the circuit shown below, 0.25 A of current flows through a 20-Ω resistor. How much voltage is needed to produce this current?

balu736 [363]

Answer:

D 5 V

Explanation:

Without seeing the whole circuit it is impossible to say for certain.

However the simplest circuit would produce a value of

FV = IR = 0.25(20) = 5 v

5 0

3 years ago

In 1970, a rocket powered car called Blue Flame achieved a maximum speed of 1.00(10 km/h (278m/s).Suppose the magnitude of the c

sammy [17]

Answer:

Distance traveled during this acceleration will be 6950 m

Explanation:

Wear have given maximum speed tat will be equal to final speed of the car v = 278 m/sec

Constant acceleration $a=5.56m/sec^2$

As the car starts initially starts from rest so initial velocity of the car u = 0 m/sec

From third equation of motion $v^2=u^2+2as$

Putting all values in equation

$278^2=0^2+2\times 5.56\times s$

s = 6950 m

So distance traveled during this acceleration will be 6950 m

3 0

3 years ago

Why is density used to determine the identity of matter?

dedylja [7]

a substance's density is the same at a certain pressure and temperature, and the density of one substance is usually different than another substance.

4 0

2 years ago

Boyle's Law mainly involves _______.

goblinko [34]

Your answer is B, gases

6 0

3 years ago

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago