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lilavasa [31]
4 years ago
10

What is the main responsibility of scientists when enviermental laws are proposed?

Chemistry
1 answer:
LekaFEV [45]4 years ago
4 0
To prove that they were tested to see if they were correct or not.
I think:/
You might be interested in
Write the oxidation and reduction half reactions;
luda_lava [24]

Answer:

a)

Fe^{2+}⇒Fe^{3+}+e^-

Br_2+2e^-⇒2Br^-

b)

Mg⇒Mg^{2+}+2e^-

Cr^{3+}+e^-⇒Cr^{3+}

Explanation:

A)

Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:

Fe^{2+}⇒Fe^{3+}+e^-

The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:

Br_2+2e^-⇒2Br^-

b)

Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:

Mg⇒Mg^{2+}+2e^-

Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:

Cr^{3+}+e^-⇒Cr^{3+}

3 0
4 years ago
Read 2 more answers
The independent variable has control and affects the
Stella [2.4K]

Answer:

The independent variable is the condition that you change in an experiment. It is the variable you control.

Explanation:

It is called independent because its value does not depend on and is not affected by the state of any other variable in the experiment. Sometimes you may hear this variable called the "controlled variable" because it is the one that is changed.

3 0
3 years ago
Read 2 more answers
How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56
neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of [A_0] is decomposed. So,

\frac {[A_t]}{[A_0]} = 0.0156

Thus,

\frac {[A_t]}{[A_0]}=e^{-k\times t}

0.0156=e^{-k\times t}

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}

t = 6\times t_{1/2}

<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>

6 0
4 years ago
Which type of radioactive decay does not cause change in the atomic mass of the product
Lady bird [3.3K]

Answer:

Gamma decay/radiation

Explanation:

Gamma radiation has no mass and no electrical charge which means no change in the atomic number or mass number when gamma rays are emitted.

4 0
3 years ago
What is one chemical reaction that begins with petroleum as a starting material
emmasim [6.3K]

Answer:

Thermal decomposition or cracking

Explanation:

Petroleum is a mixture of hydrocarbons which are usually formed naturally. Petroleum undergo a host of chemical reactions. One of such is thermal decomposition or cracking.

Cracking is used in the petroleum industry to covert heavy fractions to more useful lighter ones.

When petroleum is subjected to high temperature and pressure, and in the presence of catalyst, the long chain type of petroleum will decompose into more useful smaller and lighter molecules.

Example is given below:

                  C₁₅H₃₂ → C₈H₁₈ + C₃H₆ + 2C₂H₄

6 0
4 years ago
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