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Ksju [112]
3 years ago
8

ON a meter stick, one centimeter is generally the___ a)length of the whole stick b)length of the one hundred sticks c)distance b

etween two of the numbered lines d)distance between two of the small, unnumbered lines NEED HELP ASAP
Chemistry
1 answer:
Elena-2011 [213]3 years ago
7 0

c)distance between two of the numbered lines

Explanation:

On a meter stick, a centimeter is usually the distance between two of the numbered lines. A meter stick is a device for measuring the length of an object.

  • A hundred centimeters makes up a meter and this is the number of divisions on a meter stick.
  • Each distance between numbered vertical lines is a centimeter apart.
  • On a meter stick that is 1m long, there are 100cm.

Learn more:

lengths brainly.com/question/4687435

#learnwithBrainly

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Decarboxylation is a chemical reaction that removes a carboxyl group and releases carbon dioxide (CO2). Usually, decarboxylation refers to a reaction of carboxylic acids, removing a carbon atom from a carbon chain.

Explanation:

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A student placed equal volumes of honey and of water in two, identical, open dishes, and left them at room temperature for 8 hou
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Answer: Honey has a much lower vapor pressure than pure water has. So, pure water evaporates at a much higher rate.

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Does gas have a definite denisty?​
sergeinik [125]

Answer:

<h2><em>no</em></h2>

Explanation:

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In hydrate ratio does the mass of salt come first or does the mass of water
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6 0
3 years ago
At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---&gt; +H2(g) + Br2 (g). If the initial partial pressures o
Damm [24]

Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

4 0
3 years ago
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