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Andreyy89
3 years ago
11

What is 7.09 x 10^5 - 6.10 x 10^4 in scientific notation

Chemistry
1 answer:
Rus_ich [418]3 years ago
3 0
If you want to subtract them in scientific notation, first you need to make both exponents equal. One of them is raised to 5 and the other is raised to 4. 

7.09 x 10^{5}

6.10 x 10^{4} 

To make the first one raised to four, all you need to do is move the decimal to the right. The new notation will then be:

70.9 x 10^{4}

Now all you need to do is line up the decimals like you would normally and then subtract. Also copy the 10^{4}

70.9 x 10^{4}
-6.10 x 10^{4} 
=
64.8 x 10^{4}

Then we to make it in standard form, just move the decimal to the left and add 1 to the exponent. 

6.48 x 10^{5}
 
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A solution contains 0.140 mol KCl in 2.100 L water. What is the concentration of KCl in g/L?
viva [34]
<h3><u>Answer</u>;</h3>

≈ 4.95 g/L

<h3><u>Explanation;</u></h3>

The molar mass of KCl = 74.5 g/mole

Therefore; 0.140 moles will be equivalent to ;

 = 0.140 moles × 74.5 g/mole

 = 10.43 g

Concentration in g/L

   = mass in g/volume in L

   = 10.43/2.1

   =  4.9667

<h3>   <u> ≈ 4.95 g/L</u></h3>
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1 Copy and complete using the words below:
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Answer:

The elements in__Group_ 0 of the Periodic Table are called the_noble__gases. They are generally __unreactive_. because they have a__full_outer shell of electrons. So they do not need to gain__lose_or share _electrons_ with other atoms.

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Which organelle is like armor for the cell?
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The answer is the cell wall
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The isotopes of an element all have the same __(atomic, mass) __number, but they have different __(atomic,mass)__numbers.
Vlad1618 [11]
<h2>The isotopes of an element all have the same __(atomic, mass) __number, but they have different __(atomic,mass)__numbers.</h2>

Explanation:

The isotopes of an element all have the same __atomic  number  __, but they have different __mass __numbers.

The isotopes have same atomic number that is :

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The isotopes have different mass number that is :

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3 years ago
Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring
Lady_Fox [76]

Answer:

The standard cell potential is 1.40 V. The correct option is the option  D (+1.40 V)

Explanation:

Oxide-reduction reactions, also called redox, involve the transfer or transfer of electrons between two or more chemical species. In these reactions two substances interact: the reducing agent and the oxidizing agent.

An oxidizing element or oxidizing agent is one that reaches a stable energy state as a result of which the oxidant is reduced and gains electrons. The oxidizing agent causes oxidation of the reducing agent generating the loss of electrons of the substance and, therefore, oxidizes in the process.

In other words, the oxidizing agent is that chemical species that in a redox process accepts electrons released by the reducing agent and, therefore, is reduced in said process. The oxidizing agent is reduced because, upon receiving electrons from the reducing agent, a decrease in the value of the charge or oxidation number of one of the atoms of the oxidizing agent is induced .

Electrochemical cells, galvanic cells or batteries are called devices that are capable of transforming chemical energy originated in a spontaneous redox process into electrical energy.

The cellular potential is generally in standard conditions, that is, 1 M with respect to solute concentrations in solution and 1 atm for gases.

In this case you have the reaction:

3 Cl₂(g) + 2 Fe(s) → 6 Cl⁻(aq) + 2 Fe³⁺(aq)

In this case the following half-reactions occur:

Semi-reaction of oxidation ( an atom or group of atoms loses electrons, or increases its positive charges): Fe³⁺(aq) + 3 e- -->Fe(s); E⁰ = -0.04 V

Semi-reaction of reduction (an atom or group of atoms gains electrons, increasing its negative charges): Cl₂(g) + 2 e- --> 2 Cl-(aq); E⁰=1.36 V

In an electrochemical cell at 25°C  the potentials of the  semi-reactions are usually measured  in the sense of reduction  and generally the standard potential between both electrochemical cells will be:

E^{0} =E^{0} _{reduction} -E^{0} _{oxidation}

E⁰=1.36 V - (-0.04 V)

E⁰=1.36 V + 0.04 V

<em>E⁰=1.40 V</em>

<em><u>The standard cell potential is 1.40 V. The correct option is the option  D (+1.40 V)</u></em>

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