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lana [24]
2 years ago
5

Convert from moles to particles 6.02x1023 particles 1 mole Question 2. 2.7 moles of lithium Question 3. 1.8 moles of sodium chlo

ride​
Chemistry
1 answer:
spayn [35]2 years ago
4 0

Considering the definition of Avogadro's number:

  • the number of molecules of lithium is 1.62621×10²⁴ molecules.
  • the number of molecules of sodium chloride is 1.08414×10²⁴ molecules.

<h3>Definition of Avogadro's number</h3>

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.

<h3>Amount of molecules in this case</h3>

You can apply the following rule of three, considering the Avogadro's number:  If 1 mole of lithium contains 6.023×10²³ molecules, 2.7 moles of lithium contains how many molecules?

amount of molecules of lithium= (6.023×10²³ molecules × 2.7 moles)÷1 mole

<u><em>amount of molecules of lithium=1.62621×10²⁴ molecules</em></u>

Finally, the number of molecules of lithium is 1.62621×10²⁴ molecules.

On the other hand you can apply the following rule of three, considering the Avogadro's number:  If 1 mole of sodium chloride​ contains 6.023×10²³ molecules, 1.8 moles of  sodium chloride​ contains how many molecules?

amount of molecules of sodium chloride= (6.023×10²³ molecules × 1.8 moles)÷1 mole

<u><em>amount of molecules of sodium chloride=1.08414 ×10²⁴ molecules</em></u>

Finally, the number of molecules of sodium chloride is 1.08414×10²⁴ molecules.

Learn more about Avogadro's Number:

brainly.com/question/11907018

brainly.com/question/1445383

brainly.com/question/1528951

#SPJ1

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When a 3.00 grams sample of a compound containing only c, h, and o was completely burned, 1.17 grams of h2o and 2.87 grams of co
SVEN [57.7K]

Answer:- CH_2O_2

Solution:- From given masses of carbon dioxide and water we could calculate the moles that helps to calculate the moles of C and H.

Molar mass of carbon dioxide = 44 gram per mol

molar mass of water = 18.02 gram per mol

From given info, combustion of compound gives 1.17 grams of water and 2.87 grams of carbon dioxide. Let's calculate the moles of these:

1.17gH_2O(\frac{1mol}{18.02g})

= 0.0649molH_2O

Similarly, 2.87gCO_2(\frac{1mol}{44g})

= 0.0652molCO_2

One mol of water has two moles of H. So, the moles of H would be two times the moles of water as calculated above.

So, moles of H = 2* 0.0649 = 0.1298 mol

One mol of carbon dioxide contains one mol of C. So, the moles of C would be equal to the moles of carbon dioxide calculated above.

moles of C = 0.0652 mol

Let's convert the moles of H and C to grams so that we could calculate the amount of oxygen present in the sample as:

grams of H in sample = 1.008 x 0.1298 = 0.1308 g

grams of C in sample = 12*0.0652 = 0.7824 g

If we subtract the sum of the masses of C and H from sample mass then it would give as the mass of oxygen since the sample has only C, H and O.

mass of O in sample = 3.00g - (0.1308 g + 0.7824 g)

= 3.00 g - 0.9132 g

= 2.0868 g

Let's convert these grams of oxygen to moles on dividing by it's atomic mass as:

2.0868gO(\frac{1mol}{15.999g})

= 0.130 mol O

Now, we have the moles of all the three atoms and we know that an empirical formula is the simplest whole number ratio of the moles of atoms. So, let's calculate the ratio. For this, we divide the moles of each by the least one of them.Looking at the moles, the least value is for carbon. So, let's divide the moles of each by the moles of C as:

C = \frac{0.0652}{0.0652}  = 1

H = \frac{0.1298}{0.0652}  = 2

O = \frac{0.130}{0.0652}  = 2

The ratio of C, H and O is 1:2:2. So, the simplest formula of the compound is CH_2O_2 .



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g Ethanol boils at 78.4 °C with \DeltaΔHvap = 38.6 kJ/mol. A 0.200-mol sample of ethanol is heated from some colder temperature
andrezito [222]

Answer:

8.77 kilo Joules  will be the total amount of heat required for both the heating and the vaporizing.

Explanation:

Moles of ethanol of ethanol = 0.200 mol

Heat required to heat 0.200 moles of ethanol = Q = 1.05 kJ

Enthalpy of vaporization of ethanol = \Delta H_{vap}=38.6 kJ/mol

Heat required to vaporize 0.200 moles of ethanol = Q'

Q'=\Delta H_{vap}\times 0.200 mol=38.6 kJ/mol\times 0.200 mol=7.72 kJ

Total heat required to fore heating and the vaporizing :

= Q + Q' = 1.05 kJ + 7.72 kJ = 8.77 kJ

8.77 kilo Joules  will be the total amount of heat required for both the heating and the vaporizing.

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4 years ago
4. Convert the following: a. 4g mol of MgCl2 to g b. 2 lb mol of C3H8 to g c. 16 g of N2 to lb mol d. 3 lb of C2H6O to g mol
Nuetrik [128]

Answer:

a) 381.2 g

b) 39916 g

c) 0.0013 lb mol

d) 29.6 g mol

Explanation:

The molecular weight (mw) of a compound is the mass of it per mole, so it's the ratio of the mass (m) per mole (n).

a) The molecular weight of one mol is found at the periodic table. So, for Mg, mw = 24.3 g/mol, for Cl = 35.5 g/mol, so for MgCl2, mw = 24.3 + 2*35.5 = 95.3 g/mol. The g mol is the mass divided by the molecular weight:

g mol = m/mw

4 = m/95.3

m = 381.2 g

b) The pound (lb) is a unity of mass, and the lb mol is a unity of the mass divided by the molecular weight. So, by the periodic table, the molecular weight of C3H8 is 3*12 (of C) + 8*1 (of H) = 44 lb/mol.

lb mol = m/mw

2 = m/44

m = 88 lb

1 lb = 453.592 g

So, m = 88*453.592 = 39916 g

c) The molecular weight of N2 is 2*14 (of N) = 28 lb/mol.

m = 16/453.592 = 0.0353 lb

lb mol = m/mw

lb mol = 0.0353/28

lb mol = 0.0013 lb mol

d) The molecular weight is 2*12 (of C) + 6*1(of H) + 1*16(of O) = 46 g/mol

3 lb = 1360.78 g

g mol = m/mw

g mol = 1360.78/46

g mol = 29.6 g mol

6 0
3 years ago
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