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3 years ago

All organisms need a place to live that provides

Chemistry

1 answer:

olchik [2.2K]3 years ago

5 0

Answer:

Living Space get food or nutrients, water, and shelter. No matter where an organism lives, it must provide what it needs to survive.

Explanation:All organisms need a place to live, and store energy to take care of itself.

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Ferris & Mona used the ORP sensor to titrate a ferrous ammonium sulfate solution, (NH4)2Fe(SO4)2 with KMnO4 titrant.

DIA [1.3K]

Answer:

The correct option is: A. 0.168 M

Explanation:

Chemical reaction involved:

5 Fe²⁺ (aq) + MnO₄⁻ (aq) + 8 H⁺ (aq) → 5 Fe³⁺ (aq) + Mn²⁺ (aq) + 4 H₂O

Given: For MnO₄⁻ solution-

Number of moles: n₁ = 1, Volume: V₁ = 20.2 mL, Concentration: M₁ = 0.0250 M;

For Fe²⁺ solution:

Number of moles: n₂ = 5, Volume: V₂ = 15 mL, Concentration: M₂ = ?M

To find out the concentration of Fe²⁺ solution (M₂), we use the equation:

$\frac{M_{1}\times V_{1}}{n_{1}} = \frac{M_{2}\times V_{2}}{n_{2}}$

$\frac{0.0250 M\times 20.2 mL}{1} = \frac{M_{2}\times 15 mL}{5}$

$0.505 = M_{2}\times 3$

$M_{2} = \frac{0.505}{3} = 0.168M$

Therefore, the concentration or molarity of Fe²⁺ solution: M₂ = 0.168 M

8 0

4 years ago

Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:

Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

4 0

3 years ago

In this forest, trees block much of the sunlight from reaching the ground. The branches of trees grow up and out to reach as muc

lions [1.4K]

This example represents the challenge of survival of the fittest. In this situation, the trees have a distinct advantage due to their above average height. This puts them in the best position to gain the resources that they need to survive, most notably, the sun. The smaller plants, however, do not have this advantage, and lose out to the trees.

8 0

3 years ago

What is SEAgel help plz

DanielleElmas [232]

SEAgel (Safe Emulsion Agar gel) is one of a class of high-tech foam materials known as aerogels.

5 0

3 years ago

Read 2 more answers

I need chemistry help text me at (951) 897-8325 PLEASE

user100 [1]

Just post the question on here

7 0

3 years ago