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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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Which electromagnetic wave has the longest wavelength?
tia_tia [17]
<span>from longest wavelength to shortest: radio waves, microwaves, </span>infrared<span>, </span>optical<span>, </span>ultraviolet<span>, X-rays, and gamma-rays</span>
6 0
3 years ago
If a sample of HF gas at 694.9 mmHg has a volume of 3.463 Land the volume is changed to 5.887 L, then what will be the new press
VARVARA [1.3K]

Answer:

\large \boxed{\text{381.7 mmHg}}

Explanation:

Data:

p₁ = 694.9 mmHg; V₁ = 3.463 L

p₂ = ?;                     V₂ = 5.887 L

Calculation:

\begin{array}{rcl}p_{1}V_{1} & = & p_{2}V_{2}\\\text{648.9 mmHg} \times \text{3.463 L} & = & p_{2} \times\text{5.887 L}\\\text{2247.1 mmHg} & = & 5.887p_{2}\\p_{2} & = & \dfrac{\text{2247.1 mmHg}}{5.887}\\\\& = &\textbf{381.7 mmHg}\\\end{array}\\\text{The new pressure of the gas is $\large \boxed{\textbf{381.7 mmHg}}$}

4 0
3 years ago
For the following reaction, 9.30 grams of glucose (C6H12O6) are allowed to react with 13.8 grams of oxygen gas. glucose (C6H12O6
amid [387]

Answer:

13.7 g of CO₂

Limiting reactant:  C₆H₁₂O₆

3.81 g of O₂

Explanation:

We convert the mass of the reactants to moles, in order to find out the limiting reactant and the excess reagent

9.30 g / 180 g/mol = 0.052 moles of glucose

13.8 g / 32 g/mol = 0.431 moles of oxygen

The equation is:  C₆H₁₂O₆(s) + 6O₂ (g) → 6CO₂ (g) + 6H₂O (l)

Ratio is 1:6. Let's consider this rule of three:

1 mol of glucose reacts with 6 moles of oxygen

Then, 0.052 moles of glucose must react with (0.052 . 6) /1 = 0.312 moles

We have 0.431 moles of oxygen and we only need 0.312 moles. This means that an amount of oxygen still remains after the reaction is complete:

0.431 - 0.312 = 0.119 moles. We convert the moles to mass:

0.119 mol . 32 g / 1mol = 3.81 g

In conclussion, the limiting reactant is the glucose.

6 moles of oxygen react with 1 mol of glucose

0.431 moles of O₂ will react with (0.431 . 1) /6 = 0.072 moles of glucose

We only have 0.052 moles, so it is ok to say, that glucose is the limiting cause we do not have enough glucose.

Let's verify, the maximum amount of carbon dioxide that can be formed:

1 mol of glucose can produce 6 moles of CO₂

Therefore 0.052 moles of glucose will produce (0.052 . 6) /1 = 0.312 moles

We convert the moles to mass → 0.312 mol . 44 g /1 mol = 13.7 g

6 0
3 years ago
What is the volume of 2.5 moles of nitrogen gas (N2)<br>at standard temperature and pressure (STP)?​
gregori [183]

Answer:

1 mole of gas = 22.4L

2.5 moles of gas takes up = ( 22.4 L/ 1 mole ) x 2.5 mole

= 56 L

Explanation:

Please mark brainliest and have a great day!

8 0
3 years ago
PLSS CAN ANYONE HELP ME..​
rodikova [14]

Answer:

there is two mixture homogeneous and heterogeneous mixture

ok

7 0
2 years ago
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