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3 years ago

Using the following phase diagram, the area labeled A is in what phase?

1 answer:

8 0

<span>Using the following phase diagram, the area labeled A is in what phase?</span>Using the following phase diagram, the area labeled A is in the liquid phase.

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Construct a model to show the movement of matter and energy from plants into other organisms. Show how mass and energy are conse

ad-work [718]

Plants obtain energy from the sun during photosynthesis and pass it along the food chain.

<h3>What is a model?</h3>

A model is a representation of reality. We use models to understand what the real situation of things is like.

Let us recall that plants are the producers in the ecosystem. Energy flow from plants to animals. Plants obtain energy via the reaction 6CO2 + 6H2O --->C6H12O6 + ATP. This energy is passed along the food chain as organisms feed on each other.

Learn more about models: brainly.com/question/14281845

3 0

2 years ago

Which of the following atoms would gain two electrons to fill its valence energy level?

Harlamova29_29 [7]

The atom that would gain two electrons to fill its valence energy level is S(sulfur)

This is because s (sulfur) is in atomic number 16 with 2.8.6 of [Ne] 3s^2 2p^4 electronic configuration. This implies that sulfur has 6 valence electron and therefore it require two electron to fill its valence energy level and obtain 18 rule electrons.

5 0

3 years ago

Read 2 more answers

Balance the following redox reaction occurring in an acidic solution. The coefficient of Mn2+(aq) is given. Enter the coefficien

Vadim26 [7]

Answer:

_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)

Explanation:

Step 1:

The unbalanced equation:

AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

Step 2:

Balancing the equation.

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

The above equation can be balanced as follow:

There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)

There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)

Now the equation is balanced

7 0

3 years ago

Describe an acid or base in terms of its relationship to protons

motikmotik

Acids are donors of protons (H+) and bases are acceptors of protons.
For example:
1) hydrochloric acid (HCl) in reaction with water give one proton to water and become chloride anion (Cl-).
2) ammonia (NH3) Is base, in reaction with water accepts one protone and become ammonium cation (NH4+).

4 0

2 years ago

What volume, in milliliters, of a 0.997 M KOH solution is needed to neutralize 30.0 mL of 0.0400 M HCl?

deff fn [24]

Answer:

1.2 mL

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated:

HCl + KOH —> KCl + H₂O

From the balanced equation,

Mole ratio of the acid, HCl (nₐ) = 1

Mole ratio of base, KOH (n₆) = 1

Finally, we shall determine the volume of the base, KOH needed to neutralize the acid, HCl as follow:

Molarity of base, KOH (M₆) = 0.997 M

Volume of acid, HCl (Vₐ) = 30 mL

Molarity of acid, HCl (Mₐ) = 0.0400 M

Volume of base, KOH (V₆) =?

MₐVₐ / M₆V₆ = nₐ/n₆

0.04 × 30 / 0.997 × V₆ = 1/1

1.2 / 0.997 × V₆ = 1

Cross multiply

0.997 × V₆ = 1.2

Divide both side by 0.997

V₆ = 1.2 / 0.997

V₆ = 1.2 mL

Thus, the volume of the base, KOH needed to neutralize the acid is 1.2 mL.

7 0

3 years ago