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3 years ago

After the hare population increased, what happened to the lynx population? Why?

Chemistry

1 answer:

Westkost [7]3 years ago

5 0

The lynx population increased as well because hare are the prey and the lynx are the predators.

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Explain the structure of atom in Boron

mr Goodwill [35]

Answer: Diagram of the nuclear composition, electron configuration, chemical data, and valence orbitals of an atom of boron-11 (atomic number: 5), the most common isotope of the element boron. The nucleus consists of 5 protons (red) and 6 neutrons (orange). Five electrons (white) occupy available electron shells (rings).

Explanation:

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3 years ago

El etanol combustiona con el oxigeno produciendo dióxido de carbono gaseoso y agua: En un proceso se hacen reaccionar 450.2 g de

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Answer:

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3 years ago

ITS DUE TODAY PLSSS HELP John went to the everglades in 1994 why did heh hehehe heheh eheheheheh ehehehe frdee points ezzzzz

sp2606 [1]

Answer:

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Explanation:

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3 years ago

A scientist produces zinc iodide (zni2). (a) calculate the minimum mass of zinc that needs to be added to 0.500 g of iodine so t

harina [27]

The minimum mass of Zinc, Zn that needs to be added to 0.500 g of iodine so that the iodine fully reacts is 0.128 g

<h3>Balanced equation </h3>

Zn + I₂ —> ZnI₂

Molar mass of Zn = 65 g/mol

Mass of Zn from the balanced equation = 1 × 65 = 65 g

Molar mass of I₂ = 127 × 2 = 254 g/mol

Mass of I₂ from the balanced equation = 1 254 = 254 g

SUMMARY

From the balanced equation above,

254 g of I₂ required 65 g of Zn

<h3>How to determine the mass of Zn needed </h3>

From the balanced equation above,

254 g of I₂ required 65 g of Zn

Therefore,

0.5 g of I₂ will require = (0.5 × 65) / 254 = 0.128 g of Zn

Thus, the minimum mass of Zn required is 0.128 g

Learn more about stoichiometry:

brainly.com/question/14735801

4 0

3 years ago

A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo

atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

Weight = $0.09 \times 16.04 + 0.91 \times 29$

= 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

$\frac{800}{27.8336}$

= 28.74 kmol/hr

Now, applying component balance as follows.

$0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}$

$F_{p}$ = 43.11 kmol/hr

$F_{A} + F_{B} = F_{p}$

$F_{B}$ = 43.11 - 28.74

= 14.37 kmol/hr

So, mass flow rate of pure (B), is $F_{B}$ = $14.37 \times 29$

= 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

Molecular weight of product stream = $Mol. weight \times 43.11 kmol/hr$

= $0.06 \times 16.04 + 0.94 \times 29$

= 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of $O_{2}$ into the product stream is as follows.

$0.21 \times 0.94 \times 43.11$

= $8.5099 kmol/hr \times 329 g/mol$

= 272.31 kg/hr

Therefore, calculate the mass % of $O_{2}$ into the stream as follows.

$\frac{272.31}{1216.67} \times 100$

= 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of $O_{2}$ in the product gas is 22.38%.

4 0

3 years ago