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Margaret [11]
3 years ago
13

Diana and Kinsey are put in charge of choosing a mascot for their basketball team. There are fifteen players on the team, but Di

ana and Kinsey survey only the five players on the starting lineup about their favorite mascot.
Which best describes this error in data collection?

The data are inadequate.
The data are false.
The data were recorded incorrectly.
The data cannot be reproduced.
Chemistry
1 answer:
castortr0y [4]3 years ago
7 0

Answer: the data are inadequate .

Explanation:

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Heather just drank 40.0 grams of water (H20). How many moles of water did she just drink? O a. 2.22 moles b. 45 moles O c. 40.0
Ksju [112]

Answer:

This question begins with something, you should know: molar mass from water is aproximately 18 g/m, so if 18 grams of water are contained in 1 mole, the 40 grams occuped 2.22 moles. As you see, opcion a is the best!

Explanation:

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Gerald's science teacher mixed room temperature samples of hydrochloric acid and sodium hydroxide in a large beaker. The solutio
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 <span>HCl is a strong acid and NaOH is a strong base, so mixing them together will produce a lot of heat. The products are salt or Sodium chloride (NaCl) and water (H2O). Since one is an acid and one is a base they will neutralize each other. I hope this helps</span>
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PLEASE HELP ME ASAP! CHEMISTRY TUTOR<br><br> SEE ATTACHED
Masteriza [31]

Answer:

\large \boxed{\text{-827.4 kJ}}

Explanation:

We have three equations:

1. 2H₂S(g)            + O₂(g)   ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic)  + O₂(g)   ⟶ SO₂(g);                                 ∆H = -296.8 kJ

3. PbO(s)             + H₂S(g) ⟶ PbS(s)               + SO₂(g);    ∆H =  -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s)            + 3O₂(g) ⟶2PbO(s)             + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.  

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ  

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.  

7. 2S(s, rhombic)  + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s)  + <u>2H₂O(g</u>)  ⟶ 2PbO(s) + <u>2H₂S(g</u>);  ∆H =  208.6 kJ

6. <u>2H₂S(g)</u> + O₂(g)        ⟶ <u>2S(s</u>)     + <u>2H₂O(g)</u> ; ∆H = -442.4 kJ

<u>7</u><u>. </u><u>2S(s)</u><u>      + 2O₂(g)      ⟶ 2SO₂(g);                   ∆H = -593.6 kJ </u>

4 . 2PbS(s) + 3O₂(g)      ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}

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C, potassium. Hope this helps.

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Tomtit [17]
B) The elements become less reactive.
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