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Ilya [14]
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
Chemistry
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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How many milliliters of ammonium sulfate solution having a concentration of 0.218 M are needed to react completely with 62.6 ml
BartSMP [9]

Answer:

330 mL of (NH₄)₂SO₄ are needed

Explanation:

First of all, we determine the reaction:

(NH₄)₂SO₄  +  2NaOH →  2NH₃  +  2H₂O  +  Na₂SO₄

We determine the moles of base:

(First, we convert the volume from mL to L) → 62.6 mL . 1L/1000 mL = 0.0626L

Molarity . volume (L) = 2.31 mol/L . 0.0626 L = 0.144 moles

Ratio is 2:1. Therefore we make a rule of three:

2 moles of hydroxide react with 1 mol of sulfate

Then, 0.144 moles of NaOH must react with (0.144 .1) /2 = 0.072 moles

If we want to determine the volume → Moles / Molarity

0.072 mol / 0.218 mol/L = 0.330 L

We convert from L to mL → 0.330L . 1000 mL/1L = 330 mL

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3 years ago
How many grams of CaCl 2 are in .48 moles ?
Llana [10]

Answer:

We assume you are converting between moles CaCl2 and gram. You can view more details on each measurement unit: molecular weight of CaCl2 or grams This compound is also known as Calcium Chloride. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles CaCl2, or 110.984 grams.

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3 years ago
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Which of the following molecules are considered inorganic?
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The answer is D water
7 0
2 years ago
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A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi
Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of Na = 5.0 g

Mass of Cl_2 = 10.0 g

Molar mass of Na = 23 g/mol

Molar mass of Cl_2 = 71 g/mol

First we have to calculate the moles of Na and Cl_2.

\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}

\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol

and,

\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}

\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

2Na+Cl_2\rightarrow 2NaCl

From the balanced reaction we conclude that

As, 2 mole of Na react with 1 mole of Cl_2

So, 0.217 moles of Na react with \frac{0.217}{2}=0.108 moles of Cl_2

From this we conclude that, Cl_2 is an excess reagent because the given moles are greater than the required moles and Na is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaCl

From the reaction, we conclude that

As, 2 mole of Na react to give 2 mole of NaCl

So, 0.217 mole of HCl react to give 0.217 mole of NaCl

Now we have to calculate the mass of NaCl

\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl

Molar mass of NaCl = 58.5 g/mole

\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g

Now we have to calculate the percent yield of this reaction.

Percent yield = \frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = \frac{5.24g}{12.7g}\times 100

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

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3 years ago
Which of the followings are true 1,3-BPG? A. It regulates Hb B. It contains a high-energy bond C. It contains two ester bonds D.
weeeeeb [17]

Answer:

D. It contains a phosphate with higher phosphoryl transfer potential than ATP

Explanation:

1,3-Bisphosphoglycerate contains a phosphate group that has high phosphoryl transfer potential than ATP (they can transfer the phosphoryl group to ATP). Other high phosphoryl transfer potential groups include :Creatine kinase and phosphoenolpyruvate.

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