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Yuki888 [10]
3 years ago
7

Which elements will bond ionically with barium such that the formula would be written as BaX2? A) Nitrogen, chlorine, and sodium

Eliminate B) Chlorine, iodine, and fluorine C) Hydrogen, bromine and aluminum D) oxygen and sulfur
Physics
2 answers:
erma4kov [3.2K]3 years ago
7 0

The answer would be:

B. Chlorine, iodine and Fluorine

Barium has 2 valence electrons. To satisfy the BaX₂ , this would mean that Barium will need to give one of each of its electrons. The elements that need 1 electron would be those that have 7 valence electrons to complete the octet. These elements would fall in group 7 or halogens. Chlorine, iodine and fluorine are all in Group 7, so this would be the best choice.

Zielflug [23.3K]3 years ago
5 0
Barium with highelectronegative element that is cl iodine
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A piece of wood has mass of 250g and volume of 400cm cubed. what is its density​
Zigmanuir [339]

Answer:

0.625 \: g {cm}^{ - 3}

Explanation:

density \:  =  \frac{mass}{volume} \\  \\  =  \frac{250}{400}   \\  \\  = 0.625 \: g {cm}^{ - 3}

5 0
2 years ago
In one contest at the county fair, a spring-loaded plunger launches a ball at a speed of 3.2m/s from one corner of a smooth, fla
lara31 [8.8K]

Answer:

Explanation:

Given

Speed of ball u=3.2\ m/s

Plane is inclined at an angle 20^{\circ}

To win the Game we need to hit the target at x=2.4\ m away

Launch angle of ball \theta

Motion of ball can be considered in two planes i.e. Vertical to the plane and horizontal to the plane

So Net acceleration in vertical plane is g\sin 20

Range of Projectile is given by

R=\frac{u^2\sin 2\theta }{g}

for R=2.4\ m

2.4=\frac{3.2^2\times sin 2\theta }{g\sin 20}

\sin 2\theta =\frac{2.4\times 9.8\times \sin 20}{3.2^2}

\sin 2\theta =0.7855

2\theta =51.77

\theta =25.88^{\circ}

so ball must be launched at an angle of 25.88^{\circ}

4 0
3 years ago
A 2.8-kg cart is rolling along a frictionless, horizontal track towards a 1.2-kg cart that is held initially at rest. The carts
agasfer [191]

Answer with Explanation:

We are given that

Mass of one cart,m_1=2.8 kg

Mass of second cart,m_2=1.2 kg

Initial velocity of one cart,u_1=4.6m/s

Initial velocity of second cart,u_2=-2.7 m/s

a.Total momentum,P=m_1u_1+m_2u_2=2.8(4.6)+1.2(-2.7)

P=9.64 kgm/s

b.Velocity of second cart,v_2=0

According to law of conservation of momentum

Initial momentum=Final momentum

9.64=2.8v_1+1.2\times 0

v_1=\frac{9.64}{2.8}

v_1=3.44m/s

8 0
3 years ago
The coefficients of friction between the load and the flatbed trailershown are μs = 0.40 and μk = 0.30. Knowing that the speed o
SOVA2 [1]

Answer:

50.97 m

Explanation:

m = Mass of truck

\mu_s = Coefficient of static friction = 0.4

v = Final velocity = 0

u = Initial velocity = 72 km/h = \dfrac{72}{3.6}=20\ \text{m/s}

s = Displacement

Force applied

F=ma

Frictional force

f=\mu_s mg

Now these forces act opposite to each other so are equal. This is valid for the case when the load does not slide

ma=\mu_s mg\\\Rightarrow a=\mu_s g\\\Rightarrow a=0.4\times 9.81\\\Rightarrow a=3.924\ \text{m/s}^2

Since the obect will be decelerating the acceleration will be -3.924\ \text{m/s}^2

From the kinematic equations we have

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-20^2}{2\times -3.924}\\\Rightarrow s=50.97\ \text{m}

So, the minimum distance at which the car will stop without making the load shift is 50.97 m.

5 0
3 years ago
According to Coulomb's law, the force of attraction or repulsion between two charges is
Maru [420]

Answer:

B is the answer

Explanation:

thank you I hope it helps you

6 0
3 years ago
Read 2 more answers
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