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nexus9112 [7]
3 years ago
15

If the toxic quantity is 1.5 gg of ethylene glycol per 1000 gg of body mass, what percentage of ethylene glycol is fatal

Chemistry
1 answer:
Elina [12.6K]3 years ago
4 0

Answer:

The percentage of ethylene glycol that is fatal is 0.15 %

Explanation:

If the toxic quantity of ethylene glycol in a 1kg or 1000 g body weight is 1.5 g then the percentage of ethylene glycol that is fatal is

\frac{1.5}{1000} ˣ 100 = 0.15%

Hence, the percentage of ethylene glycol that is toxic for any body weight is 0.15%. This percentage is very important in various aspects of science including drug discovery and food production/processing

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I need help please ​
Alex17521 [72]

Answer: If you think about it, B. would be the most reasonable answer with the given factors.

4 0
3 years ago
How many moles are 21.67 L of NH4CI?<br> Type your answer...
NNADVOKAT [17]

Answer:

0.967mole

Explanation:

Given parameters:

Volume of NH₄Cl  = 21.67L

Unknown:

Number of moles  = ?

Solution:

We assume that the volume was taken at standard temperature and pressure,

 Then;

  Number of moles  = \frac{volume }{22.4L}  

 Number of moles  = \frac{21.67}{22.4}   = 0.967mole

6 0
3 years ago
Formula unit of Magnesium oxide and Calcium bicarbonate and aluminum carbonate plzzzz​
Fudgin [204]

Magnesium oxide : MgO

Calcium bicarbonate:  Ca(HCO3)2

aluminum carbonate: Al2(CO3)3 or C3Al2O9

5 0
3 years ago
What is the product of the unbalanced equation below?
Stolb23 [73]

Answer:

D

Explanation:

The answer is D. I'm not sure that it is a solid. I don't think it is a ppte, which is the only way it can be a true solid. It is ionic if the reaction is taking place in water and there is someway to start the reaction. Be that as it may, the internal balace numbers of the chemical produced is the only possible answer. The balanced eq;uatioon is

2Al + 3Br2 ==> 2AlBr3

3 0
3 years ago
If 842 grams of sodium hydroxide reacts with 750.0 grams of aluminum, how many grams of aluminum hydroxide should theoretically
Phantasy [73]

548.55 grams of aluminum hydroxide should theoretically form.

Explanation:

Balanced equation for the reaction:

3 NaOH + Al ⇒ Al(OH)3 +3 Na

DATA GIVEN:

mass of NaOH = 842 grams, atomic mass =39.9 grams/mole

mass of Al = 750 grams, atomic mass = 26.9 grams/mole

aluminum hydroxide theoretical yield = ?

Moles of NaOH reacted

number of moles = \frac{mass}{atomic mass of 1 mole}

putting the values in the equation

NaOH = \frac{842}{39.9}

           = 21.1 MOLES OF NaOH

Al = \frac{750}{26.9}

   = 27.8 moles

from the equation

 from 3 moles of NaOH 1 mole of Al(OH)3 is produced

21.1 moles of NaOH will react to give x moles of Al(OH)3

\frac{1}{3} = \frac{x}{21.1}

7.03 moles of Al(OH)3 is formed.

and

1 mole of Al(OH)3 is formed from 1 mole of Al in the reaction

so, 27.8 Moles will react to give give 27.8 moles of Al(OH)3 limiting reagent of the given reaction is NaOH

mass of Al(OH)3 =7.03 x 78 (atomic mass of Al(OH)3)

          = 548.55 grams

theoretical  yield from the given data is 548.55 grams

3 0
3 years ago
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