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Marina86 [1]
3 years ago
8

Four forces are exerted on each of the two objects shown below:

Chemistry
1 answer:
zzz [600]3 years ago
4 0

Answer:

A: Object A will move in the up direction. But not object B

Explanation:

It will move in the up direction because object A has an unbalanced force. For the up direction there, it is 6 newtons and for the bottom direction, it is 2 newtons. For the right and left direction, both of them are 3 newtons. So object A will move in the up direction because it has the most newtons(which is 6).

For Object B, it is unbalanced too. But object B will move to the right because it is 2 newtons there and there is one newton on the left. Furthermore, the up and down directions are 3 newtons. So they are balanced and will cancel out. Therefore, object A will move in the right direction because it has the most newtons(which is 3).

So the final answer is object A. Hope it helped!

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? Answer the question below. Type your response in the space provided. What volume of a 2.5 M stock solution of acetic acid (HC2
Novay_Z [31]
Data: 
M_{concentrated} = 2.5\:mol
V_{concentrated} = ?
M_{dilute} = 0.50\:mol
V_{dilute} = 100\:mL\to0.100\:L
<span>
Formula: Dilution Calculations

</span>M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}
<span>
Solving:

</span>
M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}
2.5 * V_{concentrated} = 0.50 * 0.100
2.5V_{concentrated} = 0.05
V_{concentrated} =  \frac{0.05}{2.5}
\boxed{\boxed{V_{concentrated} = 0.02\:L\:or\:20\:mL}} \end{array}}\qquad\quad\checkmark<span>







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3 years ago
The ph of a 0.0100 m solution of the sodium salt of a weak acid is 11.00. what is the ka of the acid?
Vitek1552 [10]
The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
     pOH = 14 - pH = 14 - 11.00 = 3.00
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     [OH-] = 10^-pOH = 10^-3 = x

Considering the sodium salt NaA in water, we have the equation
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Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
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     Initial             0.0100            0       0
     Change        -x                    +x     +x
     Equilibrium    0.0100-x         x       x

We can now calculate the Kb for A-:
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     Ka = Kw / Kb 
           = 1.00x10^-14 / 1.00x10^-4
           = 1.00x10^-10
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