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DedPeter [7]
3 years ago
5

From goal line to goal line, a football field is 300 ft long. If a player catches the ball while standing on one goal line, runs

to the other goal line, and scores a touchdown, how many meters did he run?
Chemistry
1 answer:
KATRIN_1 [288]3 years ago
3 0

Answer:

The player ran 91.44m.

Explanation:

The problem gives you the total distance between goal line to goal line in feet, and the answer must be given in meters, so you should convert the distance the player run from ft to m, because the player run the same distance from goal line to goal line to scores the touchdown.

So, you should apply the following conversion factor:

300ft*\frac{0.3048m}{1ft}= 91.44m

The player ran 91.44m.

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Max is helping his parents move to a new house. He picks up one box and is able to carry it into the house. He tries to pick up
Helen [10]

Answer:

The Weight

Explanation:

Even know it has the same volume does not mean it has the same mass and weighs the same with the items inside the box. Most likely, that the Items inside the box are heavier then the first box Max moved.

4 0
3 years ago
Substances that dissolve in water nd give up OH-ions are considered what
just olya [345]

Your answer would be an Arrhenius base

Hope this helps

7 0
1 year ago
What is the molarity of a solution which contains 0.80 mol of HCl in 250mL of<br> solution?
mariarad [96]

Answer:

3.2M HCl Solution

Explanation:

Molarity = moles of solute / volume of solution expressed in liters

moles of solute = 0.80 moles HCl

volume of solution = 250 ml = 0.250 Liter

Molarity (M) = 0.80 moles HCl / 0.250 Liters = 3.2M HCl Solution

6 0
2 years ago
Yield for this reaction?<br> Reaction: N2(g) + 3 H2(g) → 2 NH3 (g)
andriy [413]

Answer:

Yes, yield.

Explanation:

N2(g) + 3 H2(g) → 2 NH3 (g) balanced equation

First, find limiting reactant:  

Moles H2 = 1.83 g x 1 mole/2 g = 0.915 moles H2

Moles N2 = 9.84 g N2 x 1 mole/28 g = 0.351 moles N2

The mole ratio of H2: N2 is 3:1, so H2 is limiting (0.915 is less than 3 x 0.351)

Theoretical yield of NH3 = 0.915 mol H2 x 2 mol NH3/3 mol H2 = 0.61 moles NH3

6 0
2 years ago
Complete the following single replacement reaction. If they don’t react, just write “NR”
Kipish [7]

Here we have to complete the given single replacement reactions.

The replacement reactions are-

1) Fe (s) + CuCl₂ (aq) → FeCl₂ (aq) + Cu (s)

2) Cu (s) + FeCl₂ (aq) → NA

3) K (s) + NiBr₂ (aq) → NA

4) Ni (s) + KBr (aq) → NiBr₂ (aq) + K (s)

5) Zn (s) + Ca(NO₃)₂ (aq) → Zn(NO₃)₂ (aq)  + Ca (s)

6) Ca (s) + Zn(NO₃)₂ (aq) → NA

The replacement reactions can be explained in light of the redox potential.

The standard reduction potential of the half cells involved in these reactions are:

Fe²⁺ + 2e → Fe (E° = -0.441V); Cu²⁺ + 2e → Cu (E° = 0.674V)

Ni²⁺ + 2e → Ni (E° = -0.23V); Zn²⁺ + 2e → Zn (E° = -0.763V)

We know the half cell reactions in which the standard reduction potentials are positive are allowed.

1) The reaction is possible as Cu²⁺/Cu and Fe/Fe²⁺ standard reduction potentials are positive.

2) The reaction is not possible as Cu/Cu²⁺ and Fe²⁺/Fe standard reduction potentials are negative.

3) The reaction is not possible as Ni²⁺/Ni standard reduction potential is negative.

4) The reaction is possible as Ni/Ni²⁺ standard reduction potential is positive.

5) The reaction is possible as Zn/Zn²⁺ standard reduction potential is positive.

6) The reaction is possible as Zn²⁺/Zn standard reduction potential is negative.

4 0
2 years ago
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