All categories

2 years ago

How do ethical guidelines make science safer ?

Chemistry

2 answers:

alina1380 [7]2 years ago

6 0

Explanation:

ethical guidelines are made to be implemented together to make any type of ethical judgment.

ethical guidelines make science safer because

It provides restrictions to the scientist, they can't claim their results, which may be inaccurate or would cause any harm to the surroundings and society.

As per the Ethical guidelines, researchers are allowed to follow certain rules while conducting out research, so that their work does not cause any harm to the environment and participants.

Thepotemich [5.8K]2 years ago

5 0

Well, for one thing, they ensure that a scientist can't claim a certain result, which might be biased and thus inaccurate, because he or she is being paid by, say, big business interests. For another, they ensure that medicines or treatments can't be made available to the public without thorough testing, replication, and peer review.

You might be interested in

Why is copper (i) sulfate Cu2So4^-2? not copper (ii)?

const2013 [10]

Copper (I) oxidation state is 1 Cu2So4
copper (II) oxidation state is +2 CuSo4
copper (i) also give up one electron so you need two of them to react with the sulfate ion (which has charge of -2)
and also all metallic ions have an multiple oxidation levels corresponding to the number of electrons they can exchange or loose
Hope this helps

5 0

2 years ago

Read 2 more answers

The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO3)

cestrela7 [59]

Answer:

a. The limiting reactant is $NaHCO_{3}$

b. 0.73 g of carbon dioxide are formed.

c. The grams of excess reactant that do not participate in the reaction are 0333 g.

Explanation:

You know the following reaction:

$3NaHCO_{3} +H_{3} C_{6} H_{5} O_{7}$ ⇒ $3CO_{2} +3H_{2} O+Na_{3} C_{6} H_{5} O_{7}$

First, you determine the molar mass of each compound. For that you must take into account the atomic mass of each element:

Na: 23
H: 1
C: 12
O: 16

To determine the molar mass of each compound, you multiply the most atomic of each element present in the molecule by the sub-index that appears after each number, which indicates the present amount of each element in the compound:

$NaHCO_{3}$ :23+1+12+16*3=84 g/mol
$H_{3} C_{6} HO_{7}$ :1*3+12*6+1*5+16*7= 192 g/mol
$CO_{2}$ :12+16*2= 44 g/mol
$H_{2} O$ :1*2+16= 18 g/mol
$Na_{3} C_{6} H_{5} O_{7}$ : 23*3+12*6+1*5+16*7= 258 g/mol

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you know that 3 moles of $NaHCO_{3}$ react with 1 mole of $H_{3} C_{6} HO_{7}$ Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 192 g

You know that in a certain experiment you have 1.40 g of sodium bicarbonate and 1.40 g of citric acid. To determine the limiting reagent apply a rule of three simple as follows:

If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of sodium bicarbonate react with 1.4 grams of citric acid?

$grams of sodium bicarbonate= \frac{1.4 g*252 g}{192 g}$

grams of sodium bicarbonate= 1.8375 g

But to perform the experiment you have only 1.4 g of sodium bicarbonate. So the limiting reagent is sodium bicarbonate.

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first.

Now, by stoichiometry of the reaction, you know that 3 moles of $NaHCO_{3}$ react with 3 mole of $CO_{2}$ . Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 132 g

You make a simple rule of three: if 252 g of sodium bicarbonate form 132 g of carbon dioxide per stochetry, how many grams will form 1.4 g of sodium bicarbonate?

$grams of carbon dioxide =\frac{1.4 g * 132 g}{252 g}$

grams of carbon dioxide= 0.73 g

Then, 0.73 g of carbon dioxide are formed.

To know how much citric acid will react you apply a rule of three, taking into account as in the previous cases the stoichiometry of the reaction: If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of citric acid will they react with 1.4 g of sodium bicarbonate?

$grams of citric acid=\frac{1.4 g * 192 g}{252 g}$

grams of citric acid= 1.067 g

But you have 1.4 g of citric acid. That means that the grams you have minus the grams that react will be the grams that remain in excess and do not participate in the reaction:

grams of excess reactant=1.4 g - 1.067 g

grams of excess reactant=0.333 g

So the grams of excess reactant that do not participate in the reaction are 0333 g.

3 0

2 years ago

Please help!

Len [333]

It will slowly rotten and turn brown

7 0

3 years ago

What is true for two pieces of iron at the same temperature?

Ber [7]

D. The average kinetic energy of their particles is the same.

5 0

2 years ago

Read 2 more answers

Calculate the number of moles of each compound, given the number of molecules.

anastassius [24]

Answer:

1. $2.46\times 10^{21}$ molecules of CO₂

2. 10⁴ molecules of H₂O

3. 8.75×10³² molecules of C₆H₁₂O₆

Explanation:

1. $2.46\times 10^{21}$ molecules of CO₂

2. 10⁴ molecules of H₂O

3. 8.75×10³² molecules of C₆H₁₂O₆

5 0

2 years ago