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zmey [24]
3 years ago
9

What’s the independent variable, dependent variable, experimental variable and constant?Scenario #1: Traci thinks that instrumen

tal music played in the office will increase the productivity
of workers at the office. She creates two groups of 50 workers each and assigns each group the
same task in this case, they are asked to staple a set of papers). Group A has instrumental music
playing while they work. Group B does not have any music playing for them. After an hour, Traci
counts how many stacks of papers each group has made. Group A made 2,113 stacks. Group B
made 1,587 stacks.
Chemistry
1 answer:
saw5 [17]3 years ago
3 0

Answer:

independent variable- instrumental music

dependent variable- Number of stacks of papers made

experimental variable- Group A has instrumental music

playing

constant- same task,

Explanation:

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Which of the following would be the most likely adaptation of a whale population to tolerate a rise in ocean temperatures?
GREYUIT [131]
It is better for the whales to go where the whales have the ability to dive longer and deeper to find cooler water.
5 0
3 years ago
What is the Ka of a weak acid (HA) if the initial concentration of weak acid is 4.5 x 10-4 M and the pH is 6.87? (pick one)
rewona [7]

Answer:

Ka = 4.04 \times 10^{-11}

Explanation:

Initial concentration of weak acid = 4.5 \times 10^{-4}\ M

pH = 6.87

pH = -log[H^+]

[H^+]=10^{-pH}

[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M

HA dissociated as:

HA \leftrightharpoons H^+ + A^{-}

(0.00045 - x)    x     x

[HA] at equilibrium = (0.00045 - x) M

x = 1.35 \times 10^{-7}\ M

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

Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}

0.000000135 <<< 0.00045

Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}

5 0
3 years ago
In a study of the formation of NOx air pollution, a chamber heated to 2200°C was filled with air (0.790 atm N₂, 0.210 atm O₂). W
Irina-Kira [14]

Answer:

N₂ = 0.7515atm

O₂ = 0.1715atm

NO = 0.0770atm

Explanation:

For the reaction:

N₂(g) + O₂(g) ⇄ 2NO(g)

Where Kp is defined as:

Kp = \frac{P_{NO}^2}{P_{N_2}P_{O_2}}}

Pressures in equilibrium are:

N₂ = 0.790atm - X

O₂ = 0.210atm - X

NO = 2X

Replacing in Kp:

0.0460 = [2X]² / [0.790atm - X] [0.210atm - X]

0.0460 = 4X² / 0.1659 - X + X²

0.0460X² - 0.0460X + 7.6314x10⁻³ = 4X²

-3.954X² - 0.0460X + 7.6314x10⁻³ = 0

Solving for X:

X = - 0.050 → False answer. There is no negative concentrations.

X = <em>0.0385 atm</em> → Right answer.

Replacing for pressures in equilibrium:

N₂ = 0.790atm - X = <em>0.7515atm</em>

O₂ = 0.210atm - X = <em>0.1715atm</em>

NO = 2X = <em>0.0770atm</em>

3 0
4 years ago
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Answer:

second law of acceleration

Explanation:

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3 years ago
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Smallest particle of a substance that can retain its properties &amp; is
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Answer:

Molecules

Explanation:

8 0
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