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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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The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.320 with the floor. If
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Answer:  29.50 m

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f=μ*N the friction force is equal to coefficient of static friction  multiply the normal force (m*g).

f=m.a=μ*N= m*a= μ*m*g= m*a

then

a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2

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as follows:

x= vo*t- (a/2)* t^2

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5 0
3 years ago
Help me please I can't get the final step​
inna [77]

Answer:

\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

\displaystyle A=\frac{3}{2}B^mC^n

and the dimensions of each variable is:

A=L^2T^2

B=LT^{-1}

C=LT^2

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n

Operating:

L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)

L^2T^2=L^{m+m}T^{-m+2n}

Equating the exponents:

m+n=2

-m+2n=2

Adding both equations:

3n=4

Solving:

n=4/3

m=2-4/3=2/3

Answer:

\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}

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3 years ago
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A mass m = 550 g is hung from a spring with spring constant k = 2.8 N/m and set into oscillation at time t = 0. A second, identi
riadik2000 [5.3K]

Answer:

The second system must be set in motion t=0.70s seconds later

Explanation:

The oscillation time, T, for a mass, m, attached to spring with Hooke's constant, k, is:

T=2\pi\sqrt(\frac{m}{k} )

One oscillation takes T secondes, and that is equivalent to a 2π phase. Then, a difference phase of π/2=2π/4, is equivalent to a time t=T/4.

If the phase difference π/2 of the second system relative to the first oscillator. The second system must be set in motion t=\frac{\pi}{2}\sqrt(\frac{m}{k})=\frac{\pi}{2}\sqrt(\frac{0.55}{2.8}= 0.70s) seconds later

6 0
4 years ago
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