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nataly862011 [7]
3 years ago
8

Which type of eye wash station must be checked and flushed most often

Chemistry
2 answers:
Marina CMI [18]3 years ago
6 0

stand up eye wash station


Elanso [62]3 years ago
5 0

Answer:

<u><em>The answer is</em></u>: <u>When a foreign substance, usually irritating, penetrates the eyes, and for people with sensitive eyes.</u>

<u />

Explanation:

An eyewash liquid, or eyewash liquid system is a liquid, <em>usually a saline solution,</em> used as an emergency or safety mechanism <u><em>in laboratories and industrial plants</em></u><em> as an aid to eye rinsing, s</em> when a foreign substance, usually irritating, penetrates in them.

<em>These eye drops can be beneficial </em>for people with sensitive eyes <em>and can provide relief from the side effects of painful sensation. </em>

<u><em>The answer is</em></u>: <u>When a foreign substance, usually irritating, penetrates the eyes, and for people with sensitive eyes.</u>

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When FeC13 is ignited in an atmosphere of pure oxygen, this reaction takes place. 4FeCl3(sJ 30lgJ ~ 2F~0 (sJ 6Cl2(gJ If 3.00 mol
oksano4ka [1.4K]

Answer : The reagent present in excess and remains unreacted is, O_2

Solution : Given,

Moles of FeCl_3 = 3.00 mole

Moles of O_2 = 2.00 mole

Excess reagent : It is defined as the reactants not completely used up in the reaction.

Limiting reagent : It is defined as the reactants completely used up in the reaction.

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2FeCl_3(s)+O_2(g)\rightarrow 2FeO(s)+3Cl_2(g)

From the balanced reaction we conclude that

As, 2 moles of FeCl_3 react with 1 mole of O_2

So, 3.00 moles of FeCl_3 react with \frac{3.00}{2}=1.5 moles of O_2

From this we conclude that, O_2 is an excess reagent because the given moles are greater than the required moles and FeCl_3 is a limiting reagent and it limits the formation of product.

Hence, the reagent present in excess and remains unreacted is, O_2

4 0
3 years ago
In the reaction C + O2 → CO2, 18 g of carbon react with oxygen to produce 72 g of carbon dioxide. What mass of oxygen would be n
artcher [175]
Stoichiomety:

1 moles of C + 1 mol of O2 = 1 mol of CO2

multiply each # of moles times the atomic molar mass of the compund to find the relation is weights

Atomic or molar weights:

C: 12 g/mol
O2: 2 * 16 g/mol = 32 g/mol
CO2 = 12 g/mol + 2* 16 g/mol = 44 g/mol

Stoichiometry:

12 g of C react with 32 g of O2 to produce 44 g of CO2

Then 18 g of C will react with: 18 * 32/ 12 g of Oxygen = 48 g of Oxygen

And the result will be 12 g of C + 48 g of O2 = 60 g of CO2.

You cannot obtain 72 g of CO2 from 18 g of C.

May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.
3 0
4 years ago
Read 2 more answers
Calculate the molality and van’t Hoff factor (i) for the following aqueous solutions:
MissTica

The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

<h3>What is the value of Van t Hoff factor?</h3>

For most non-electrolytes dissolved in water, the Van 't Hoff factor is essentially $ 1 $ . For most ionic compounds dissolved in water, the Van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.

<h3>Which has highest Van t Hoff factor?</h3>

The Van't Hoff factor will be highest for

   A. Sodium chloride.

   B. Magnesium chloride.

   C. Sodium phosphate.

   D. Urea.

Learn more about van't off factor here:

<h3>brainly.com/question/22047232</h3><h3 /><h3>#SPJ4</h3>

3 0
2 years ago
Potential energy is what
____ [38]

Answer:

the energy possessed by a body by its value of its position relative to others, stresses within itself, electric charge, and other factors.

Explanation:

4 0
2 years ago
A metallurgist has one alloy containing 21% aluminum and another containing 42% aluminum. how many pounds of each alloy must he
Naily [24]
Answer is: 7.8 lb of 21% aluminum and 33.2 ib of <span>42% aluminum.</span>

ω₁<span> = 21% ÷ 100% = 0.21.
ω</span>₂<span> = 42% ÷ 100% = 0.42.
ω</span>₃<span> = 38% ÷ 100% = 0.38.
</span>m₁ = ?.

m₂<span> = ?.
</span>m₃ = m₁ + m₂<span>.
</span>m₃ = 41 pounds.

m₁ = 41 lb - m₂<span>.
ω</span>₁ · m₁ + ω₂ ·m₂ = ω₃ · m₃.

0.21 · (41 lb - m₂) + 0.42 · m₂ = 0.38 · 41 lb.

8.61 lb - 0.21m₂ + 0.42m₂ = 15.58 lb.

0.21m₂ = 6.97 lb.

m₂ = 6.97 lb ÷ 0.21.

m₂ = 33.2 lb.

m₁ = 41 lb - 33.2 lb.

m₁<span> = 7.8 lb.</span>



8 0
3 years ago
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