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Mice21 [21]
3 years ago
7

Min is asked to balance the equation below.

Physics
2 answers:
Minchanka [31]3 years ago
6 0
It is incorrect, because the identity of the original product has changed. Ca3(OH)2 does not exist! It is no longer calcium hydroxide. To balance an equation, you must manipulate the coefficients, a.k.a. the big numbers that go before reactants or products. Subscripts, the little numbers inside the reactants or products, cannot be changed without completely changing the substance.
Wewaii [24]3 years ago
6 0

Ca3P2 + H2O → Ca(OH)2 + PH3

Min’s result: Ca3P2 + H2O → Ca3(OH)2 + PH3

Which best describes Min’s method for balancing the chemical equation?

It is correct, because now there are three Ca atoms on each side of the equation.

It is incorrect, because the identity of the original product has changed.

It is correct, because Ca3(OH)2 creates a more stable compound.

It is incorrect, because the subscript “2” should be added instead.

Answer: B

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Need help solving this question.
MatroZZZ [7]

Answer:

See the answers below.

Explanation:

to solve this problem we must make a free body diagram, with the forces acting on the metal rod.

i)

The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.

We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.

For the summation of forces we will take the forces upwards as positive and the negative forces downwards.

ΣF = 0

-15+T-W=0\\T-W=15

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.

ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.

ΣM = 0

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7 0
2 years ago
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him
juin [17]

Answer:

The velocity of the student has after throwing the book is 0.0345 m/s.

Explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s

Angle = 31°

We need to calculate the magnitude of the velocity of the student has after throwing the book

Using conservation of momentum along horizontal  direction

m_{b}v_{b}\cos\theta= m_{c}v_{c}

v_{s}=\dfrac{m_{b}v_{b}\cos\theta}{m_{c}}

Put the value into the formula

v_{c}=\dfrac{1.25\times3.61\times\cos31}{112}

v_{c}=0.0345\ m/s

Hence, The velocity of the student has after throwing the book is 0.0345 m/s.

3 0
3 years ago
Need physics help<br> ASAP please!
Effectus [21]

Answer:

D

Explanation:

7 0
3 years ago
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