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Feliz [49]
3 years ago
14

Can you help me with this?

Physics
1 answer:
Natasha2012 [34]3 years ago
4 0

Answer:

no

Explanation:

so basically I am domb so I can not help you

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The compound PCl5 decomposes into Cl2 and PCl3. The equilibrium of PCl5(g) Cl2(g) + PCl3(g) has a Keq of 2.24 x 10-2 at 327°C. W
nordsb [41]
Take note of the reaction formula which is PCl5=Cl2+PCl3.
The Keq = [Cl2] * [PCl3] / [PCl5]=2.24*10^-2.
For the reason that the volume is 1 liter, the concentration of Cl2 will be computed through: <span>(2.24 * 10^-2) * 0.235 / 0.174 </span> = 0.0303 mol/L is the answer.
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If a magnet repels a metal object what can you say about the pole of the magnet and the pole of the object?
sleet_krkn [62]
Poles are the same D is the right answer
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I NEED THIS A.S.A.P I WILL GIVE BRAINLIEST PLEASE HELP!!
VLD [36.1K]

Explanation:

change 0.5 g to kg so 0.005kg then change 100 ml to m so 0.001m so density=mass over volume so from there you can continue

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Consider the points below. P(1, 0, 1), Q(−2, 1, 4), R(6, 2, 7) (a) Find a nonzero vector orthogonal to the plane through the poi
kozerog [31]

Answer:

a) (0, -33, 12)

b) area of the triangle : 17.55 units of area

Explanation:

<h2>a) </h2>

We know that the cross product of linearly independent vectors \vec{A} and \vec{B} gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.

Luckily for us, we know that vectors \vec{A} = \vec{P}-\vec{Q} and \vec{B} = \vec{R} - \vec{Q} are living in the plane through the points P, Q, and R, and are linearly independent.

We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).

If they weren't linearly independent, we will obtain vector zero as the result of the cross product.

So, for our problem:

\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)

\vec{B} = \vec{R} - \vec{Q} \\\\\vec{B} = (6,2,7) - (-2,1,4)\\\\\vec{B} = (6 +2,2-1,7-4)\\\\\vec{B} = (8,1,3)

\vec{A} \times  \vec{B} = (A_y B_z - B_y A_z) \  \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}

\vec{A} \times  \vec{B} = ( (-1) * 3 - 1 * (-3) ) \  \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}

\vec{A} \times  \vec{B} = ( - 3 + 3 ) \  \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}

\vec{A} \times  \vec{B} = 0 \  \hat{i} - 33 \ \hat{j} + 12 \ \hat{k}

\vec{A} \times  \vec{B} =(0, -33, 12)

<h2>B)</h2>

We know that \vec{A} and \vec{B} are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:

|\vec{A} \times  \vec{B} | = 2 * area_{triangle}

so:

\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}

\sqrt{1233} = 2 * area_{triangle}

35.114= 2 * area_{triangle}

17.55 \ units \  of \ area =  area_{triangle}

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If you are given distance and a period of time, you can calculate the speed. The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of its speed and direction of motion (e.g. 60 km/h to the north).

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