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3 years ago

Helppppp meeeeeeeeeeeeeeeeeeeeeeeeeeeee ill mark brainlist

Chemistry

2 answers:

hammer [34]3 years ago

7 0

Answer: C

Explanation:

I think the answers correct because, A says that the food chain will stay the same, but thats not true because, the frog ate most of the insects. And B says the population will decrease, but that doesn't make sense because a animal that eats grasshoppers would no longer be there. And C says that the population will increase and i think that correct because, one animal that does eat the grasshopper wouldn't be there so the grasshoppers will increase in population. And D says the population of snacks will increase but that was one of the foods the snacks ate so they would decrease because they lost one species to eat off of.

Hope This Helps!

sergey [27]3 years ago

3 0

Grasshopper population will increase since they have less predators

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4 years ago

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3 years ago

Match the following aqueous solutions with the appropriate letter from the column on the right. Assume complete dissociation of

Serhud [2]

Answer: 0.17 m $CH_3COONH_4$ : Highest freezing point

0.20 m $CoSO_4$ : Second lowest freezing point

0.18 m $MnSO_4$ : Third lowest freezing point

0.42 m ethylene glycol: Lowest freezing point

Explanation:

Depression in freezing point is a colligative property which depend upon the amount of the solute.

$\Delta T_f=i\times k_f\times m$

where,

$\Delta T_f$ = change in freezing point

i= vant hoff factor

$k_f$ = freezing point constant

m = molality

a) 0.17 m $CH_3COONH_4$

For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i =2 for $CH_3COONH_4$ , thus total concentration will be 0.34 m

b) 0.18 m $MnSO_4$

For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i = 2 for $MnSO_4$ , thus total concentration will be 0.36 m

c) 0.20 m $CoSO_4$

For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i = 2 for $CoSO_4$ , thus total concentration will be 0.40 m

d) 0.42 m ethylene glycol

For non electrolytes undergoing no dissociation, vant hoff factor is equal to 1 . Thus i = 1 for ethylene glycol, thus concentration will be 0.42 m

The more is the concentration, the highest will be depression in freezing point and thus lowest will be freezing point.

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3 years ago

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3 years ago

HELP I NEED THIS ASAP<br>ON THE IMAGE

Sunny_sXe [5.5K]