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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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do you mean polymers or organic compounds?
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A 1.3 g sample of a substance is heated from 0°C to 45°C and is found to have absorbed 45 j of heat. What is the specific heat o
wlad13 [49]

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5 0
3 years ago
When 0.49 g of a molecular compound was dissolved in 20.00 g of cyclohexane, the freezing point of the solution was lowered by 3
IRISSAK [1]

Answer: The molecular mass of this compound is 131 g/mol

Explanation:

Depression in freezing point:

\Delta T_f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}

where,

\Delta T_f = depression in freezing point  = 3.9^oC

k_f = freezing point constant  = 20.8^0C/m

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

w_2 = mass of solute = 0.49 g

w_1 = mass of solvent (cyclohexane) = 20.00 g

M_2 = molar mass of solute = ?

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

(3.9)^oC=1\times (20.8^oC/m)\times \frac{(0.49g)\times 1000}{M_2\times (20.00g)}

M_2=131g/mol

Therefore, the molar mass of solute is 131 g/mol

7 0
2 years ago
Three common gaseous compounds of nitrogen and oxygen of different elementary composition are known: (A) laughing gas containing
Ede4ka [16]

Answer:

Please find how these data prove the law of multiple proportions below

Explanation:

The law of multiple proportions was proposed by an English chemist called John Dalton. The law states that when two elements combine and to form more than one compound. The weights/masses of the second element in the two compounds, which combines with a fixed ratio of the first element, is in a simple whole number ratio.

In this question, Nitrogen is said to combine with oxygen to give three different compounds as follows:

A) laughing gas containing 63.65% nitrogen i.e. 0.6365g

This means that the mass of oxygen will be (1-0.6365) = 0.3635g

B) colorless gas containing 46.68% nitrogen i.e. 0.4668g

This means that the mass of oxygen will be 0.5332g

C) brown toxic gas containing 30.45% nitrogen i.e. 0.3045g

This means that the mass of oxygen will be 0.6955g

The ratios of oxygen in the three compounds is therefore:

0.3635: 0.5332: 0.6955

Divide this ratio by the smallest number (0.3635)

0.3635/0.3635 = 1

0.5332/0.3635 = 1.467

0.6955/0.3635 = 1.913

Multiply this ratio by 2, we have:

2: 2.9 : 3.8

Hence, the simple whole number ratio is 2:3:4.

This proves the law of multiple proportions that oxygen is in simple whole number ratio in the three different compounds.

4 0
3 years ago
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