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otez555 [7]
2 years ago
9

How many moles are in 4.3 x 1024 molecules of H20?

Chemistry
1 answer:
Bond [772]2 years ago
4 0

Answer:

\boxed {\boxed {\sf About 7.1 \ mol \ H_2O}}

Explanation:

To convert from molecules to moles, we must use Avogadro's Number: 6.022*10²³. This tells us the amount of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case, the particles are molecules of water.

\frac {6.022*10^{23} \ molecules \ H_2O} {1 \ mol \ H_2O}

Multiply by the given number of molecules.

4.3 *10^{24} \ molecules \ H_2O *\frac {6.022*10^{23} \ molecules \ H_2O} {1 \ mol \ H_2O}

Flip the fraction so the molecules of water cancel.

4.3 *10^{24} \ molecules \ H_2O *\frac {1 \ mol \ H_2O} {6.022*10^{23} \ molecules \ H_2O}

4.3 *10^{24}  *\frac {1 \ mol \ H_2O} {6.022*10^{23} }

\frac {4.3 *10^{24}\ mol \ H_2O} {6.022*10^{23} }

7.140484889 \ mol \ H_2O

The original measurement of atoms has 2 significant figures ( 4 and 3), so our answer must have the same. For the moles we calculated, that is the tenth place. The 4 in the hundredth place tells us to leave the 1.

7.1 \ mol \ H_2O

There are about 7.1 moles of water.

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The electron configuration of an element is 1s22s22p63s1 Describe what most likely happens when an atom of this element comes ne
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Answer:

They will create an ionic bond.

Explanation:

The atom with the one valence electron will lose its one, because it's a metal and metals will lose electrons to become stable. The nonmetal (with 7 valence electrons) will gain that electron, therefore creating a stable octet for the nonmetal, making the compound stable.

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2 years ago
Does anybody know how to do this if so please help !
Viktor [21]

Answer:

add x to 7 and divide by 3

Explanation:

easier formula

3 0
2 years ago
The measurement 0.0002833g has how many significant figures?
kherson [118]

Answer:

There are seven significant figures

Explanation:

There are seven different digits within the number. Three 0s, one 2, one 8, and two 3s, adding up to seven different numbers. You exclude the first 0 when the number is a decimal, leaving seven significant figures. Hope this makes sense! :)

6 0
2 years ago
How much energy is used to melt 44.33 g of solid oxygen?
Nutka1998 [239]

Answer:

Q1 = C * m * dT

Q2 = Qm * m

Qtotal = Q1 + Q2

Q1 - is amount of energy you need to apply to heat oxygen from the current temperature till you reach the melting temperature. Only if the oxygen is below to melting temperature.

C - is calorific capacity of oxygen -- better look at tables, it is a constant value

m - is the amount of oxygen, we will use moles because the other data shows moles, but could be grams, kg, etc.

dT - is the diference of temperatures between the current and the melting one. The melting temperature is constant and you can find it on tables, then (Tm - To)

Q2 is the amount of energy you have to add to melt oxygen once the oxygen has reached the melting temperature (Tm)

Qm is a constant value you could find on tables, depends on the mass of oxygen and is due to internal processes as changes in atomic distributions

If the oxygen is initially at melting temperature (melting point) you only need to know Q2, as dT = 0

I will do an example for you, but in future you should provide data of constants, it takes very long to find them in books or internet.

Data from tables

Tm =  54.36 K

C = 29.378 J/mol K this is at 25 C (or 298 K), is not really correct, you should look at its value at less than 54.36 K, but you can use it here.

Qm = 0.444 kJ/mol

Problem -- you have 44.33g of Oxygen -- Molecular weight of O2 is 32 g/mol

So you have 44.33/32 = 1.385 moles of oxygen

a) if oxygen is already at melting temperature: you only have to melt it

Qtotal = Q1 + Q2 = [0 (dT = 0) + Qm * m] = 0.444 * 1.385 = 0.615 kJ = 615 J

b) supposing an initial temperture of 50 K: now you have to heat oxygen till melting temperature and then melt it.

Q1 = C * m * dT = 29.378 * 1.385 * (54.36 - 50) = 177.442 J

Q2 = Qm * m = 615 J

Qtotal = 177.442 + 615 = 792.44 J

Explanation:

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2 years ago
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17% is the percent error

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