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2 years ago

Illustrate: The Avogadro constant is so large it is normally written in scientific notation. To

Chemistry

2 answers:

kondaur [170]2 years ago

8 0

Answer:

602200000000000000000000

Explanation:

Avogadro's number in scientific notation is:

6.022 * 10^23

Now to write it in standard form, we need to move the decimal point 23 places to the right. Since we have to move 3 places to the right to get 6022, we need to only add 23 - 3 = 20 more zeroes:

602200000000000000000000

Hope this helps!

Licemer1 [7]2 years ago

5 0

Answer:

602200000000000000000000

Explanation:

move the decimal place over 23 places to the right

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The level of water in an olympic size swimming pool (50.0 meters long, 25.0 meters wide, and about 2.00 meters deep) needs to be

MA_775_DIABLO [31]

Initial dimension of the pool:

Length (L)= 50.0 m

Width (W)= 25.0 m

Depth (D) = 2.0 m

Initial Volume of the pool (V1) = L*W*D = 50.0 * 25.0 *2.0 = 2500 m3

When the depth is lowered by 3 cm i.e. 0.03 m, the new depth becomes: 2.0 - 0.03 = 1.97 m

The new volume of the pool (V2) = 50.0*25.0*1.97 = 2462.5 m3

Volume of water to be pumped out = V1-V2 = 2500-2462.5 = 35.5 m3

Now, 1 m3 = 1000 L

therefore, 35.5 m3 == 35.5 *10^3 L

It is given that:

3.80 L of water is pumped out in 1 sec

Therefore, 35.5*10^3 L will be pumped out in:

= 1 s * 35.5*10^3 L/3.80 L = 9.34*10^3 sec

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2 years ago

Can someone answer theses problems? Only do the Even number problems,Im in a rush so please help me.

8090 [49]

The answer to the questions. All 15 of them.

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2 years ago

Which of these elements has the highest first ionization energy?

julia-pushkina [17]

Answer:

Cesium (Cs)

Explanation:

This eliment has an ionization energy of 3,8939

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2 years ago

370 cm3 of water at 80°C is mixed with 120 cm3 of water at 4°C. Calculate the final equilibrium temperature, assuming no heat is

NikAS [45]

Answer : The final temperature of the mixture is $61.4^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

$(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)$

where,

$c_1$ = $c_2$ = specific heat of water = same

$m_1$ = $m_2$ = mass of water = same

$\rho_1$ = $\rho_2$ = density of water = 1.0 g/mL

$V_1$ = volume of water at $80.0^oC$ = $370cm^3=370mL$

$V_2$ = volume of water at $4^oC$ = $120cm^3=120mL$

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of water = $80.0^oC$

$T_2$ = initial temperature of water = $4^oC$

Now put all the given values in the above formula, we get:

$(\rho_1\times V_1)\times (T_f-T_1)=-(\rho_2\times V_2)\times (T_f-T_2)$

$(1.0g/mL\times 370mL)\times (T_f-80.0)^oC=-(1.0g/mL\times 120mL)\times (T_f-4)^oC$

$T_f=61.4^oC$

Therefore, the final temperature of the mixture is $61.4^oC$

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2 years ago

An ion of a single pure element always has an oxidation number of ________. a. 1 b. magnitude equal to its atomic number c. 0 d.

harina [27]

The oxidation state of a pure element is always zero.

The oxidation state for a pure ion is equivalent to its ionic charge.

What is oxidation state?

The oxidation state or oxidation number, is the hypothetical charge of an atom if all of its bonds to different atoms were fully ionic. The oxidation state may be positive, negative or zero.

What are the rules to write the oxidation number?

1. The atoms in pure elements have oxidation number of ZERO.

2. The more-electronegative element in a binary molecular compound is assigned the number equal to the negative charge it would have as an anion. The less-electronegative element is assigned the number equal to the positive charge it would have as a cation.

3. Fluorine always has an oxidation number of -1 in compounds because it is the most electronegative element.

4. Oxygen is usually -2 unless in peroxides (then it is -1) or with fluorine.

5. Hydrogen has +1 in all compounds containing elements that are more electronegative than it, and is -1 in compounds with metals.

6. The algebraic sum of the oxidation numbers of all atoms in a neutral compound is zero.

7. The algebraic sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge of the ion.

8. Although rules 1-7 apply to covalently bonded atoms, oxidation numbers can also be assigned to atoms in ionic compound. Monoatomic ions have oxidation numbers equal to their ionic charge.

To learn more about oxidation number: brainly.com/question/3035498

#SPJ4

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11 months ago