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HACTEHA [7]
3 years ago
7

One tank of gold fish is fed the normal amount of food once a day. A second tank is fed twice a day. A third tank is fed four ti

mes a day during a six week study. The fish's weight is recorded daily. All tanks were of the same size and shape. IV: DV: Control Group:
Chemistry
1 answer:
Arturiano [62]3 years ago
4 0

The question is incomplete, the complete question is;

One tank of goldfish is feed the normal amount which is once a day, a second tank is fed twice a day, and a third tank is fed four times a day during a 6 week study. The fishes' body fat is recorded daily.

Independent Variable-

Dependent Variable-

Constants

Control Group-

Answer:

A) The amount of food the gold fish receives

B) Body fat of the gold fish

C) -Type of fish used in the study (gold fish)

Time period within which the fishes were fed (Six week period)

Shape and size of tank

D) group of gold fish fed the normal amount

Explanation:

The purpose of the study is to determined the impact of amount of feed on the body fat of gold fish. Hence, the amount of feed is the independent variable while the body fat of the feed is the dependent variable.

The control group receives the normal amount of feed (once a day). The fishes are all gold fish, fed within a six week period. All the tanks were of the same shape and size. These are the constants in the study.

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3 years ago
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13. A mixture of MgCO3 and MgCO3.3H2O has a mass of 3.883 g. After heating to drive off all the water the mass is 2.927 g. What
rjkz [21]

Answer:

63.05% of MgCO3.3H2O by mass

Explanation:

<em>of MgCO3.3H2O in the mixture?</em>

The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:

<em>Mass water:</em>

3.883g - 2.927g = 0.956g water

<em>Moles water -18.01g/mol-</em>

0.956g water * (1mol/18.01g) = 0.05308 moles H2O.

<em>Moles MgCO3.3H2O:</em>

0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =

0.01769 moles MgCO3.3H2O

<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>

0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O

<em>Mass percent:</em>

2.448g MgCO3.3H2O / 3.883g Mixture * 100 =

<h3>63.05% of MgCO3.3H2O by mass</h3>
6 0
3 years ago
8. The time period of artificial satellite in a circular orbit of radius R is T. The radius of the orbit in which time period is
Elena-2011 [213]

Explanation:

It is given that,

The time period of artificial satellite in a circular orbit of radius R is T. The relation between the time period and the radius is given by :

T^2\propto R^3

The radius of the orbit in which time period is 8T is R'. So, the relation is given by :

(\dfrac{T}{T'})^2=(\dfrac{R}{R'})^3

(\dfrac{T}{8T})^2=(\dfrac{R}{R'})^3  

\dfrac{1}{64}=(\dfrac{R}{R'})^3

R'=4\times R

So, the radius of the orbit in which time period is 8T is 4R. Hence, this is the required solution.  

4 0
3 years ago
5. A gas occupies a volume of 100.0 mL at 27.0°C and 630.0 torr. At what temperature, in degrees
DanielleElmas [232]

100 by 500 ml so if you put 50 ml in 630torr it would evaporate

Explanation:

it is more than ¹00 celcwius so it would eveporate

8 0
3 years ago
If 100 ml of a 0.5 M HCI Solution is diluted with water to 1000ml, what is the new concetration?
Vesna [10]

Answer:

0.05\ \text{M HCl}

Explanation:

V_1 = Initial volume = 100 mL

V_2 = Final volume = 1000 mL

M_1 = Initial concentration = 0.5 M

M_2 = Final concentration

We have the relation

\dfrac{M_1}{M_2}=\dfrac{V_2}{V_1}\\\Rightarrow M_2=M_1\dfrac{V_1}{V_2}\\\Rightarrow M_2=0.5\times \dfrac{100}{1000}\\\Rightarrow M_2=0.05\ \text{M HCl}

The new concentration is 0.05\ \text{M HCl}.

3 0
3 years ago
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