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3 years ago

11

Grief is conducting an experiment to see whether goldfish eat more takes or pellots. He thinks that the fish will choose the fla

kes. Every day, Geoff places both flakes and pellets in a fish tank and observes which food the fish
crone. What evidence would be best for Good to collect as he observes the fish?
the time for the fish to finish eating
the number of th that eat each type of food
themes for the shto choose what type of food to eat

1 answer:

Lyrx [107]3 years ago

5 0

Answer: The number of fish that eat each type of food

Explanation:

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A bottle of wine contains 9.81 grams of C2H5OH, dissolved in 87.5 grams of water. The final volume of the solution is 100.0 mL.

dimulka [17.4K]

Answer:

[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)

Explanation:

1. Molarity = moles solute / Volume solution in Liters

=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH

=> volume of solution (assuming density of final solution is 1.0g/ml) ...

volume solution = 9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution

Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH

2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)

From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln

= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.

3 0

3 years ago

What type of bond does hydrogen and carbon form

laila [671]

Covalent bond , hopefully this help :) explanation : .....

7 0

3 years ago

Read 2 more answers

M is a mysterious element whose molar mass is 141.0 g/mol. When combined with oxygen, the formula of the compound is M2O5. What

Alex Ar [27]

Answer:

I d k

Explanation:

7 0

3 years ago

CdF2(s)⇄Cd2+(aq)+2F−(aq)A saturated aqueous solution of CdF2 is prepared. The equilibrium in the solution is represented above.

kupik [55]

Answer:

The correct answer is option a.

Explanation:

$CdF_2(s)\rightleftharpoons Cd^{2+}(aq)+2F^-(aq)$

Equilibrium concentration cadmium ions = $[Cd^{2+}]=0.0585 M$

Equilibrium concentration fluoride ions = $[F^{-}]=0.117 M$

Molar solubility is the maximum concentration of salt present in water in ionic form beyond that no more salt will exist in its ionic form and will settle down in bottom of the solution.

The molar solubility of the solid cadmium fluoride = 0.0585 M

$CdF_2(s)\rightleftharpoons Cd^{2+}(aq)+2F^-(aq)$ ..[1]

$NaF(s)\rightleftharpoons Na^{+}(aq)+F^-(aq)$

Due to addition of sodium fluoride will increase concentration of fluoride in the solution.And due to common ion effect the equilibrium will shift in backward direction in [1], that is precipitation of more cadmium fluoride.

Hence, decrease in solubility will be observed.

8 0

3 years ago

mixas84 [53]

1) Area of one side of each cube

The area of one side of each cube can be calculated by multiplying the length times the width:

$A=L\cdot W$

where

L = length

W = width

For cube A: $L=1 cm, W=1 cm$

So the area is: $A_A = (1)(1)=1 cm^2$

For cube B: $L=2 cm, W=2 cm$

So the area is: $A_B=(2)(2)=4 cm^2$

For cube C: $L=3 cm, W=3 cm$

So the area is: $A_C=(3)(3)=9 cm^2$

2) Surface area of each cube

The surface area of a cube is given by the sum of the areas of all its faces. Since a cube has 6 identical faces, this means that the total surface area is equal to 6 times the area of one face:

$A=6 A_1$

where

$A_1$ is the area of one face of the cube

For cube A, the area of one face is $A_A=1 cm^2$

So the surface area of cube A is

$A'_A = 6A_A=6(1)=6 cm^2$

For cube B, the area of one face is $A_B=4 cm^2$

So the surface area of cube B is

$A'_B = 6A_B=6(4)=24 cm^2$

For cube C, the area of one face is $A_C=9 cm^2$

So the surface area of cube C is

$A'_C = 6A_C=6(9)=54 cm^2$

3) Volume of each cube

The volume of a cube is obtained by multiplying its length, its width and its height:

$V=L\cdot W \cdot H$

Where:

L = length

W = width

H = height

Moreover for a cube, all the sides have equal length, so $L=W=H$

So the volume can be rewritten as

$V=L^3$

For cube A: L = 1 cm

So the volume is $V_A=(1 cm)^3 = 1 cm^3$

For cube B: L = 2 cm

So the volume is $V_B=(2 cm)^3 = 8 cm^3$

For cube C: L = 3 cm

So the volume is $V_C = (3 cm)^3 = 27 cm^3$

4) Ratio surface area/volume for each cube

In this part, we have to calculate the ratio between surface area and volume of each cube:

$r=\frac{A}{V}$

where

A is the surface area

V is the volume

For cube A, we have:

$A_A = 6 cm ^2$ (surface area)

$V_A=1 cm^3$ (volume)

So the ratio for cube A is:

$r=\frac{6}{1}=6$

For cube B, we have:

$A_B = 24 cm ^2$ (surface area)

$V_B=8 cm^3$ (volume)

So the ratio for cube B is:

$r=\frac{24}{8}=3$

For cube C, we have:

$A_C = 54 cm ^2$ (surface area)

$V_C=27 cm^3$ (volume)

So the ratio for cube C is:

$r=\frac{54}{27}=2$

4 0

3 years ago