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viva [34]
3 years ago
13

Look at your gumdrop model of a water molecule. If you had several of those water molecules, and you pulled the gumdrops apart,

what other molecules could you make? Explain your answer.
Chemistry
1 answer:
Andrews [41]3 years ago
4 0
It will make a bigger molecule
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If the vapor's volume were to be incorrectly recorded as 125ml, how will this error affect the calculated molar mass of the unkn
pishuonlain [190]
Since you didn't give the actual volume (or any of the experimental values) I can only tell you how to do it. Do the calculation using the real (determined) volume of the flask. Then, re-do the calculation with v = 125ml. Take the two values and calculate % error; m = measured vol; g = guessed vol. 

<span>[mW (m) - mW (g)]/mW (m) x 100% </span>

<span>(they want % error so, if it is negative, just get rid of the sign) </span>
3 0
3 years ago
A 55.0 mL aliquot of a 1.50 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a
balu736 [363]

Answer:

0.14 M

Explanation:

To determinate the concentration of a new solution, we can use the equation below:

C1xV1 = C2xV2

Where C is the concentration, and V the volume, 1 represents the initial solution, and 2 the final one. So, first, the initial concentration is 1.50 M, the initial volume is 55.0 mL and the final volume is 278 mL

1.50x55.0 = C2x278

C2 = 0.30 M

The portion of 139 mL will be the same concentration because it wasn't diluted or evaporated. The final volume will be the volume of the initial solution plus the volume of water added, V2 = 139 + 155 = 294 mL

Then,

0.30x139 = C2x294

C2 = 0.14 M

4 0
3 years ago
Read 2 more answers
The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.80/(M · s) at 10°C. Starting with a concentrati
a_sh-v [17]

Answer : The concentration of NOBr after 95 s is, 0.013 M

Explanation :

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.80M^{-1}s^{-1}

t = time taken  = 95 s

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 0.86 M

Now put all the given values in above equation, we get:

0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)

[A] = 0.013 M

Hence, the concentration of NOBr after 95 s is, 0.013 M

4 0
3 years ago
Name the four different types of Spheres belonging to Earth and give examples of Renewable resource and the Biotic factors that
madreJ [45]

Answer:

Everything in Earth's system can be placed into one of four major subsystems: land, water, living things, or air. These four subsystems are called "spheres." Specifically, they are the "lithosphere" (land), "hydrosphere" (water), "biosphere" (living things), and "atmosphere" (air).

Explanation:

4 0
3 years ago
A- is a weak base. which equilibrium corresponds to the equilibrium constant ka for ha?
vekshin1

Answer:

K_a=\frac{[H^+][A^-]}{[HA]}

Explanation:

ka is defined as the dissociation constant of an acid. It is defined as the ratio of concentration of products to the concentration of reactants.

For the dissociation of weak acid, the chemical equation follows:

HA\rightleftharpoons H^++A^-

The equilibrium constant is defined by the equilibrium concentration of products over reactants:

K_a=\frac{[H^+][A^-]}{[HA]}

5 0
3 years ago
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