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Rainbow [258]
3 years ago
6

When preparing for work in the fume hood, be sure to gather all necessary tools, glassware, and chemicals _________ to minimize

the number of times the hood sash is raised and lowered. Work as much as possible in the _________ of the work surface to keep the area tidy and promote air flow. If you need to step away from the experiment to obtain another item, _________ the sash during this time.
Chemistry
1 answer:
andreyandreev [35.5K]3 years ago
5 0

Answer:

In advance

middle

lower

Explanation:

These are the safety precautions needed when carrying out duties in the fume hood.

When planning and preparing to work in a fume hood (a locally designed area to reduce exposure to hazardous fumes). It is advisable to make all equipment readily available at your disposal <u>in advance</u> to reduce and minimize the raising and lowering of the hood sash at intervals.

It is also pertinent to understand that working in the<u> middle </u>of the work surface helps to promote the movement of air and keeps the area neat and tidy.

However, if any case where there is a need to get a new tool or equipment during the process of working in a fume hood, it is advisable to <u>lower </u>the sash at that point in time.

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If a jar of honey sits for a long time, sugar may crystallize in the jar. Describe two possible causes for the crystallization.
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Dryness and temperature are the primary reasons but also the type of honey.
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3 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
Which of the following has zero acceleration?
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5 0
3 years ago
Read 2 more answers
A 110 g copper bowl contains 240 g of water, both at 21.0°C. A very hot 410 g copper cylinder is dropped into the water, causing
vlada-n [284]

Answer:

There is 98.76 kJ energy transfered to the water as heat.

Explanation:

<u>Step 1:</u> Data given

Mass of copper bowl = 110 grams

Mass of water = 240 grams

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Mass of the hot copper cylinder = 410 grams

8.6 grams being converted to steam

Final temperature = 100 °C

<u>Step 2:</u> Calculate the energy gained by the water:

Q = m(water)*C(water)*ΔT + m(vapor)*Lw

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⇒ with C(water) = the heat capacity of water = 4184 J/kg°C

⇒ with ΔT = the change in temperature = T2 - T1 = 100 °C - 21.0 = 79°C

⇒ with mass of vapor = 8.60 grams = 0.0086 kg

⇒ with Lw = The latent heat of vaporization (water to steam) = 22.6 *10^5 J/kg

Q = 0.24kg * 4184 J/kg°C *79°C + 0.0086 kg*22.6*10^5 J/kg

Q = 79328.64 + 19436 = 98764.64 J = 98.76 kJ

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Answer:A and D

Explanation:

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