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3 years ago

8

An unknown or changeable quantity is called a(n)...

1 answer:

alexgriva [62]3 years ago

6 0

I think the answer would be dependent variable. An unknown or changeable quantity is called a dependent variable. It <span>is what you measure in the experiment and what is affected during the experiment. Hope this answers the question. Have a nice day.</span>

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How is a mineral's effervescence measured?

Nata [24]

Dripping Acid onto it

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4 years ago

The oxidation of sulfur dioxide by oxygen to sulfur trioxide has been implicated as an important step in the formation of acid r

Igoryamba

Answer:

The value of the equilibrium constant $K_p$ at this temperature is 3.42.

Explanation:

Partial pressure of the sulfur dioxide = $p_1=0.564 atm$

Partial pressure of the oxygen gas = $p_2=0.102 atm$

Partial pressure of the sulfur trioxide = $p_3=0.333$

$2 SO_2(g) + O_2(g)\rightleftharpoons 2 SO_3(g)$

The expression of an equilibrium constant is given by :

$K_p=\frac{(p_3)^2}{(p_1)^2\times p_2}$

$K_p=\frac{(0.333 atm)^2}{(0.564 atm)\times (0.102 atm)}=3.42$

The value of the equilibrium constant $K_p$ at this temperature is 3.42.

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4 years ago

The density of water is 1.00 g/mL. If your swimming pool holds 208,000 gallons of water, what is the weight of the water in a fi

Natasha2012 [34]

Answer:

12.455≈12.5 tons

Explanation:

1 US gallon of water (gal) = 8.35 pounds of water (lb wt.)

so we divide 208,000 by 8.35 to covert to pounds

then divide the pounds by 2000 to convert to tons.

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3 years ago

One sphere has a radions of 181 cm, another has a radius of 5.01 cm. What is the difference in volume (in cubic centimeters) bet

PSYCHO15rus [73]

Answer:

$2.4*10^{7}cm^{3}$

Explanation:

First you should calculate the volume of a big sphere,so:

$V_{big}=\frac{4}{3}\pi r^{3}$

$V_{big}=\frac{4}{3}\pi (181cm)^{3}$

$V_{big}=2.4*10^{7}cm^{3}$

Then you calculate the volume of a small spehre, so:

$V_{small}=\frac{4}{3}\pi r^{3}$

$V_{small}=\frac{4}{3}\pi (5.01cm)^{3}$

$V_{small}=5.3*10^{2}cm^{3}$

Finally you subtract the two quantities:

$V_{big}-V_{small}=2.4*10^{7}cm^{3}-5.3*10^{2}cm^{3}$

$V_{big}-V_{small}=2.4*10^{7}cm^{3}$

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3 years ago

Please help! This is an exam practice but it’s really hard and I don’t really know it, any help will be appreciated!!

Olin [163]

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3 years ago