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kumpel [21]
3 years ago
5

At constant force, acceleration varies (directly, inversely) with mass. When subjected to the same amount of net external force,

a heavier object will experience (less, greater) acceleration than a lighter one.
Chemistry
2 answers:
Lady bird [3.3K]3 years ago
8 0

Answer:

The answer

Explanation:

directly

greater

lora16 [44]3 years ago
7 0

Answer:

I don't know the ans please search on the Google you will get

And don't forget to mark me as brainlest please guys and follow me back please please please please please

And I will help you tooooooooooooooooo and follow u back if you follow me

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The soldiers are issued (infrared) goggles to help them see in the dark.
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"outside visible spectrum" would be the right way to define the power of the goggles provided to the soldiers to see at night. The correct option among all the options that are given in the question is the first option or option "A". I hope the answer comes to your great help.
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What is a force, provide an example
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A force involves an interaction between two or more objects, and it causes a push or pull between the objects. ... Good examples of opposing force include drag due to interaction with an air mass and the force due to friction between two objects.

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Suppose that 4.8 L of methane at a pressure
Ghella [55]

Answer:

972.3 Torr

Explanation:

P2=P1V1/V2

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A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci
Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}

The ICE table is then shown as:

                               C_3H_6ClCO_2H_{(aq)}  \ \ \ \ \to  \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ +  \ \ \ \ H^+_{(aq)}

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

K_a  = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}

where ;

K_a = 3.02*10^{-5}

3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}

Since the value for K_a is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

3.02*10^{-5} *(1.8) = {(x)(x)}

5.436*10^{-5}= {(x^2)

x = \sqrt{5.436*10^{-5}}

x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M

Dissociated form of  4-chlorobutanoic acid = C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M

Percentage dissociated = \frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100

Percentage dissociated = \frac{7.3729*10^{-3}}{1.8 }*100

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

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Since medals form cations
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