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jeka94
3 years ago
8

Goot is conducting an experiment to see whether goldfish eat more flakes or pellets. He thinks that the fish will choose the fla

kes. Every day, Gooft places both flakes and pellets in a fish tank and observes which food the fish
choose what evidence would be best for Geoff to collect as he observes the fish?
umber of fish in the tank
the time takes for the fish to finish eating
the number of th that cat cach type of food
the time to for the fish to choose what type of food to eat
Chemistry
2 answers:
jolli1 [7]3 years ago
5 0

Answer The number of fish that eat each type of food

Explanation:

andreev551 [17]3 years ago
3 0

Answer:

the number of th that cat cach type of food

Explanation:

cuz im big brain

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For the equilibrium
Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

H_2S=0.596M

H_2=0.004 M

S_2=0.002 M

Explanation:

We are given that  for the equilibrium

2H_2S\rightleftharpoons 2H_2(g)+S_2(g)

k_c=9.0\times 10^{-8}

Temperature, T=700^{\circ}C

Initial concentration of

H_2S=0.30M

H_2=0.30 M

S_2=0.150 M

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

H_2S=0.30+2x

H_2=0.30-2x

S_2=0.150-x

At equilibrium

Equilibrium constant

K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Substitute the values

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

By solving we get

x\approx 0.148

Now, equilibrium concentration  of gases

H_2S=0.30+2(0.148)=0.596M

H_2=0.30-2(0.148)=0.004 M

S_2=0.150-0.148=0.002 M

3 0
2 years ago
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