SCIENCE QUESTION

You notice that the water in your friend's swimming pool is cloudy and that the pool walls are discolored at the water line. A q

Flauer [41]

Answer:

6,78 mL of 12,0 wt% H₂SO₄

Explanation:

The equilibrium in water is:

H₂O (l) ⇄ H⁺ (aq) + OH⁻ (aq)

The initial concentration of [H⁺] is 10⁻⁸ M and final desired concentration is [H⁺] = $10^{-7,20}$ , thus,

Thus, you need to add:

[H⁺] = $10^{-7,2} -10^{-8,0}$ = 5,31x10⁻⁸ M

The total volume of the pool is:

9,00 m × 15,0 m ×2,50 m = 337,5 m³ ≡ 337500 L

Thus, moles of H⁺ you need to add are:

5,31x10⁻⁸ M × 337500 L = 1,792x10⁻² moles of H⁺

These moles comes from

H₂SO₄ → 2H⁺ +SO₄²⁻

Thus:

1,792x10⁻² moles of H⁺ × $\frac{1H_{2}SO_4 mol}{2H^{+} mol}$ = 8,96x10⁻³ moles of H₂SO₄

These moles comes from:

8,96x10⁻³ moles of H₂SO₄ × $\frac{98,1 g}{1 mol}$ × $\frac{100 gSolution}{12 gH_{2}SO_4 }$ × $\frac{1 mL}{1,080 g}$ =

6,78 mL of 12,0wt% H₂SO₄

I hope it helps!

8 0

3 years ago

A worker is told her chances of being killed by a particular process are 1 in every 300 years. Should the worker be satisfied or

Pavlova-9 [17]

Answer:

(a) Yes, he should be worried. The Fatal accident rate (FAR) is too high according to standars of the industry. This chemical plant has a FAR of 167, where in average chemical plants the FAR is about 4.

(b) FAR=167 and Death poer person per year = 0.0033 deaths/year.

(c) The expected number of fatalities on a average chemical plant are one in 12500 years.

Explanation:

Asumming 50 weeks of work, with 40 hours/week, we have 2000 work hours a year.

In 300 years we have 600,000 hours.

With these estimations, we have (1/600,000)=1.67*10^(-6) deaths/hour.

If we have 2000 work hours a year, it is expected 0.0033 deaths/year.

$1.67*10^{-6} \frac{deaths}{hour}*2000 \frac{hours}{year}=0.0033 deaths/year$

The Fatal accident rate (FAR) can be expressed as the expected number of fatalities in 100 millions hours (10^(8) hours).

In these case we have calculated 1.67*10^(-6) deaths/hour, so we can estimate FAR as:

$FAR=1.67*10^{-6} \frac{deaths}{hour}*10^{8} hours=1.67*10^{2} =167$

A FAR of 167 is very high compared to the typical chemical plants (FAR=4), so the worker has reasons to be worried.

If we assume FAR=4, as in an average chemical plant, we expect

$4\frac{deaths}{10^{8} hour} *2000\frac{hours}{year}=8*10^{-5} \frac{deaths}{year}$

This is equivalent to say

$\frac{1}{8*10^{-5} } \frac{years}{death}=1.25*10^{4} \frac{years}{death} =12500 \, \frac{years}{death}$

The expected number of fatalities on a average chemical plant are one in 12500 years.

4 0

3 years ago

Select the correct answer from each drop-down menu. consider the substances hydrogen (h2), fluorine (f2), and hydrogen fluoride

nasty-shy [4]

The boiling point of HF is higher than the boiling point of $H_2$ , and it is higher than the boiling point of $F_2$ .

<h3>What is the boiling point?</h3>

The boiling point is the temperature at which the pressure exerted by the surroundings upon a liquid is equalled by the pressure exerted by the vapour of the liquid.

$F_2$ has weak dispersion force attractions between its molecules, whereas liquid HF has strong ionic interactions between $H^+$ and $F^-$ ions.

Only London Forces are formed - Therefore more energy is required to break the intermolecular forces in HF than in the other hydrogen halides and so HF has a higher boiling point.

$H_2$ and $F_2$ will only have intra-molecular attractions and there will be no hydrogen bonds present in them. As a result, their boiling point will be lower.

Hence, the boiling point of HF is higher than the boiling point of $H_2$ , and it is higher than the boiling point of $F_2$ .

Learn more about the boiling point:

brainly.com/question/25777663

#SPJ1

7 0

1 year ago

_ an electron moved around the nucleus in circular paths

kifflom [539]

Answer: although it is convenient to think of the electron moving around the nucleus along circular paths

Explanation:

7 0

2 years ago

graduated cylinder is filled with water to a volume of 6.2 ML. an irregular shaped plastic object weighing 1.2 g is placed in th

Aleks04 [339]

Answer:

A. Density of object = 0.86 g/mL

B. The object will float in the water.

Explanation:

The following data were obtained from the question:

Volume of water = 6.2 mL

Mass (m) of object = 1.2 g

Volume of water + Object = 7.59 mL

Density of object =?

Density of water = 1 g/mL

Next, we shall determine the volume of the object. This can be obtained as follow:

Volume of water = 6.2 mL

Volume of water + Object = 7.59 mL

Volume of object =?

Volume of object = (Volume of water + Object) – (Volume of water)

Volume of object = 7.59 – 6.2

Volume of object = 1.39 mL

Therefore, the volume of the object is 1.39 mL

A. Determination of the density of the object.

Mass (m) of object = 1.2 g

Volume (V) of object = 1.39 mL

Density (D) of object =?

Density = mass /volume

Density = 1.2/1.39

Density of object = 0.86 g/mL

B. Determination of whether the object will float or sink.

Density of object = 0.86 g/mL

Density of water = 1 g/mL

From the above, we can see that the density of water is greater than that of the object. This implies that the object is lighter than water. Therefore, the object will float in the water.

8 0

3 years ago