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2 years ago

7

for every 10 yards on a football field there I boldly marked line labeled with the amount of yards there are two parallel sideli

nes on each side of the field if the 50 yards line is perpendicular to one of the sidelines what is its relationship to the other sideline

1 answer:

RUDIKE [14]2 years ago

3 0

It is perpendicular to both as the sidelines are parallel

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In Part III, the phenolphthalein indicator is used to monitor the equilibrium shifts of the ammonia/ammonium ion system. The phe

AnnZ [28]

The pink color in the solution fades. Some of the colored indicator ion converts to the colorless indicator molecule.

<h3>Explanation</h3>

What's the initial color of the solution?

$\text{NH}_4\text{Cl}$ is a salt soluble in water. $\text{NH}_4\text{Cl}$ dissociates into ions completely when dissolved.

$\text{NH}_4\text{Cl} \; (aq)\to {\text{NH}_4}^{+} \; (aq) +{\text{Cl}}^{-} \; (aq)$ .

The first test tube used to contain $\text{NH}_4\text{OH}$ . $\text{NH}_4\text{OH}$ is a weak base that dissociates partially in water.

$\text{NH}_4\text{OH} \; (aq) \rightleftharpoons {\text{NH}_4}^{+} \;(aq)+ {\text{OH}}^{-} \; (aq)$ .

There's also an equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ ions.

${\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)$ .

$\text{OH}^{-}$ ions from $\text{NH}_4\text{OH}$ will shift the equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ to the right and reduce the amount of ${\text{H}_3\text{O}}^{+}$ in the solution.

The indicator equilibrium will shift to the right to produce more ${\text{H}_3\text{O}}^{+}$ ions along with the colored indicator ions. The solution will show a pink color.

What's the color of the solution after adding NH₄Cl?

Adding $\text{NH}_4\text{Cl}$ will add to the concentration of ${\text{NH}_4}^{+}$ ions in the solution. Some of the ${\text{NH}_4}^{+}$ ions will combine with $\text{OH}^{-}$ ions to produce $\text{NH}_4\text{OH}$ .

The equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ ions will shift to the left to produce more of both ions.

${\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)$

The indicator equilibrium will shift to the left as the concentration of ${\text{H}_3\text{O}}^{+}$ increases. There will be less colored ions and more colorless molecules in the test tube. The pink color will fade.

7 0

2 years ago

Which of the following causes a physical change in the shape of an object?

babunello [35]

Answer:

B

Explanation:

Heating a piece of iron until it glows.

3 0

2 years ago

What volume (in mL) of a 0.200 MHNO3 solution is required to completely react with 27.6 mL of a 0.100 MNa2CO3 solution according

ladessa [460]

Answer:

There is 27.6 mL of a 0.200 M HNO3 solution required

Explanation:

<u>Step 1: </u>The balanced equation is:

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

This means for 1 mole Na2CO3 consumed, there is consumed 2 mole of HNO3 and there is produced 2 moles of NaNO3, 1 mole of CO2 and 1 mole of H2O

<u>Step 2: </u>Calculating moles of Na2CO3

moles of Na2CO3 =volume of Na2CO3 * Molarity of Na2CO3

moles of Na2CO3 = 27.6 *10^-3 * 0.1 M = 0.00276 moles

<u>Step 3: </u>Calculating moles of HNO3

In the balanced equation, we can see that for 1 mole of Na2CO3 consumed, there are consumed 2 moles of HNO3.

So for 0.00276 moles consumed of Na2CO3, there are consumed 0.00552 moles of HNO3.

This means 0.00276 moles of the base Na2CO3 would react with 0.00552 moles of the acid HNO3

<u>Step 4: </u>Calculating the volume of HNO3

volume of HNO3 = moles of HNO3 / Molarity of HNO3

volume of HNO3 = 0.00552 moles / 0.200 M = 0.0276 L

0.0276 L = 27.6 ml

There is 27.6 mL of a 0.200 M HNO3 solution required

4 0

2 years ago

How many grams of aluminum chloride are produced when 5.96 grams of aluminum are reacted with excess chlorine gas? Start with a

Vadim26 [7]

Answer:

29.47 g of AlCl₃.

Explanation:

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

2Al + 3Cl₂ —> 2AlCl₃

Next, we shall determine the mass of Al that reacted and the mass of AlCl₃ produced from the balanced equation. This can be obtained as follow:

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 2 × 27 = 54 g

Molar mass of AlCl₃ = 27 + (35.5× 3)

= 27 + 106.5

= 133.5 g/mol

Mass of AlCl₃ from the balanced equation = 2 × 133.5 = 267 g

SUMMARY:

From the balanced equation above,

54 g of Al reacted to produce 267 g of AlCl₃.

Finally, we shall determine the mass of AlCl₃ produced by the reaction of 5.96 g of Al. This can be obtained as follow:

From the balanced equation above,

54 g of Al reacted to produce 267 g of AlCl₃.

Therefore, 5.96 g of Al will react to produce = (5.96 × 267)/54 = 29.47 g of AlCl₃.

Thus, 29.47 g of AlCl₃ were obtained from the reaction.

3 0

2 years ago

I NEED HELP ANSWERING THESE QUESTIONS FOR 50 POINTS!!!!!

Semmy [17]

The answers are true, true, false, true, and false.

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2 years ago

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