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Alexandra [31]
3 years ago
13

What is the iodide ion concentration in a solution if the addition of an excess of 0.100 m pb(no3)2 to 42.9 ml of the solution p

roduces 861.8 mg of pbi2?
Physics
1 answer:
Katen [24]3 years ago
3 0
Moles Pbl2 =  0.8628 g  :  461.01 g/mol   = 0.001871

moles I = 2 x  0.001871 = 0.003742

[I-] = 0.003742/ 0.0429

= 0.0872 M
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In order to be accepted, a scientific theory must be
bija089 [108]

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The answer is the 1st one

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The first PCS technology used a form of time division multiplexing called ____.​ a. ​Time Division Multiple Access (TDMA) b. ​Gl
salantis [7]

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A. Time Division multiple Access (TDMA)

Explanation:

Time-division multiple access (TDMA) is a channel access method for shared-medium networks

It allows several users to share the same frequency channel by dividing the signal into different time slots. The users transmit in rapid succession, one after the other, each using its own time slot.

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Please help I'm stuck on this question ​
Afina-wow [57]

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increase

decrease

Explanation:

using formula

Vt=mg/6πηr

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8 0
2 years ago
A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc
Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

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The distance traveled by car after Δt (seconds) at v_c = 23m/s speed is

s_c = \Delta t v_c = 23\Delta t

The distance traveled by the motorcycle after Δt (seconds) at m_m = 23 m/s speed and acceleration of a = 8 m/s2 is

s_m = \Delta t v_m + a\Delta t^2/2

s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

s_m = s_c + 58

23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58

4 \Delta t^2 = 58

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\Delta t = \sqrt{14.5} = 3.807s

(b)

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5 0
3 years ago
Determine the value of the resultant and its location from O.<br>see attach image.​
xxTIMURxx [149]

Answer:

Explanation:

In the x direction the force will be

½(-w₀)L/2 = -¼w₀L  

acting ⅔(L/2) = L/3 below the x axis.

In the y direction the force will be

½(-w₀)L + ½w₀L/2 = -¼w₀L  

the magnitude of the resultant will be

F = w₀L  √((-¼)² + (-¼)²) = w₀L√⅛

in the direction

θ = arctan(-¼w₀L / -¼w₀L) = 225°

to find the distance, we balance moments

(w₀L√⅛)[d] = ½(w₀)L[⅔L] + ¼w₀L[⅔L/2] - ¼w₀L[L - ⅓L/2]

     (√⅛)[d] = ½         [⅔L] + ¼      [⅔L/2] - ¼      [L - ⅓L/2]

     (√⅛)[d] = ½[⅔L] + ¼[⅔L/2] - ¼[L - ⅓L/2]

     (√⅛)[d] =      ⅓L  +    ⅟₁₂L     -  ¼L + ⅟₂₄L  

     (√⅛)[d] = 5L/24

               d = 5L/24 / (√⅛)

               d = 5√⅛L/3

8 0
3 years ago
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