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iVinArrow [24]
3 years ago
14

During the process of succession,

Chemistry
2 answers:
bagirrra123 [75]3 years ago
7 0
What is the question is it a fill in the blank multiple choice?
hjlf3 years ago
7 0

Answer:

producers typically enter a developing ecosystem before consumers.

Explanation:

Ecological succession is the process by which an ecosystem undergoes a series of changes as communities of organisms change their environment and new communities of organisms move in to the ecosystem. During the process of succession, producers typically enter a developing ecosystem before consumers. The producers modify the environment and makes it more suitable for the consumers.

(i had this on study island BTW)

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In Part Two, you will apply the engineering design process to design your own skate park ramp.
grandymaker [24]

Answer: The constraints could be lack of materials, funds, or time. Another constraint could be size or shape.

Explanation:

E :)

7 0
1 year ago
Read 2 more answers
What mass of Fe(OH)3 is produced when 35 mL of 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M
Zina [86]

Answer:

0.35 g.

Explanation:

We'll begin by calculating the number of mole of Fe(NO3)3 in 35 mL of 0.250 M Fe(NO3)3 solution.

This is illustrated below:

Molarity of Fe(NO3)3 = 0.250 M

Volume = 35 mL = 35/1000 = 0.035 L

Mole of Fe(NO3)3 =?

Molarity = mole /Volume

0.250 = mole of Fe(NO3)3 / 0.035

Cross multiply

Mole of Fe(NO3)3 = 0.25 x 0.035

Mole of Fe(NO3)3 = 8.75×10¯³ mole

Next, we shall determine the number of mole of KOH in 55 mL of 0.180 M

KOH solution. This is illustrated below:

Molarity of KOH = 0.180 M

Volume = 55 mL = 55/1000 = 0.055 L

Mole of KOH =.?

Molarity = mole /Volume

0.180 = mole of KOH /0.055

Cross multiply

Mole of KOH = 0.180 x 0.055

Mole of KOH = 9.9×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

3KOH + Fe(NO3)3 —> Fe(OH)3 + 3KNO3

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3 to produce 1 mole of Fe(OH)3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3.

Therefore, 9.9×10¯³ mole of KOH will react with = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(NO3)3.

From the above illustration, we can see that only 3.3×10¯³ mole out of 8.75×10¯³ mole of Fe(NO3)3 given is needed to react completely with 9.9×10¯³ mole of KOH.

Therefore, KOH is the limiting reactant and Fe(NO3)3 is the excess reactant.

Next, we shall determine the number of mole of Fe(OH)3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of Fe(OH)3 as all of it is consumed in the reaction.

The limiting reactant is KOH and the mole of Fe(OH)3 produce can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted to produce 1 mole of Fe(OH)3.

Therefore, 9.9×10¯³ mole of KOH will react to produce = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(OH)3.

Finally, we shall convert 3.3×10¯³ mole of Fe(OH)3 to grams. This can be obtained as follow:

Molar mass of Fe(OH)3 = 56 + 3(16 + 1) = 56 + 3(17) = 107 g/mol

Mole of Fe(OH)3 = 3.3×10¯³ mole

Mass of Fe(OH)3 =?

Mole = mass /Molar mass

3.3×10¯³ = Mass of Fe(OH)3 / 107

Cross multiply

Mass of Fe(OH)3 = 3.3×10¯³ x 107

Mass of Fe(OH)3 = 0.3531 ≈ 0.35 g.

Therefore, 0.35 g of Fe(OH)3 was produced from the reaction.

8 0
2 years ago
If you start with 1.000 gram sample of the isotope how much time would pass before you have just 0.100 grams of the isotope left
Vadim26 [7]

Answer:

So, you're dealing with a sample of cobalt-60. You know that cobalt-60 has a nuclear half-life of

5.30

years, and are interested in finding how many grams of the sample would remain after

1.00

year and

10.0

years, respectively.

A radioactive isotope's half-life tells you how much time is needed for an initial sample to be halved.

If you start with an initial sample

A

0

, then you can say that you will be left with

A

0

2

→

after one half-life passes;

A

0

2

⋅

1

2

=

A

0

4

→

after two half-lives pass;

A

0

4

⋅

1

2

=

A

0

8

→

after three half-lives pass;

A

0

8

⋅

1

2

=

A

0

16

→

after four half-lives pass;

⋮

Explanation:

now i know the answer

6 0
2 years ago
Which of the following measurements is equal to 2.3 dL?
Nonamiya [84]
What are the following measurements?
6 0
3 years ago
A tank at is filled with of sulfur tetrafluoride gas and of sulfur hexafluoride gas. You can assume both gases behave as ideal g
dangina [55]

The question is incomplete, the complete question is:

A 7.00 L tank at 21.4^oC is filled with 5.43 g of sulfur hexafluoride gas and 14.2 g of sulfur tetrafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas. Round each of your answers to significant digits.

<u>Answer:</u> The mole fraction of sulfur hexafluoride is 0.221 and that of sulfur tetrafluoride is 0.779

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

  • <u>For sulfur hexafluoride:</u>

Given mass of sulfur hexafluoride = 5.43 g

Molar mass of sulfur hexafluoride = 146.06 g/mol

Putting values in equation 1, we get:

\text{Moles of sulfur hexafluoride}=\frac{5.43g}{146.06g/mol}=0.0372mol

  • <u>For sulfur tetrafluoride:</u>

Given mass of sulfur tetrafluoride = 14.2 g

Molar mass of sulfur tetrafluoride = 108.07 g/mol

Putting values in equation 1, we get:

\text{Moles of sulfur tetrafluoride }=\frac{14.2g}{108.07g/mol}=0.1314mol

Total moles of gas in the tank = [0.0372+ 0.1314] mol = 0.1686 mol

Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:

\chi_A=\frac{n_A}{n_A+n_B} .....(2)

where n is the number of moles

Putting values in equation 2, we get:

\chi_{SF_6}=\frac{0.0372}{0.1686}=0.221

\chi_{SF_4}=\frac{0.1314}{0.1686}=0.779

Hence, the mole fraction of sulfur hexafluoride is 0.221 and that of sulfur tetrafluoride is 0.779

7 0
3 years ago
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