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evablogger [386]
3 years ago
10

Molar mass of propan-1-ol

Chemistry
2 answers:
Lady_Fox [76]3 years ago
8 0

Answer:

60g/mol.

Explanation:

<u>TO FIND :-</u>

  • Molar mass of Propan-1-ol

<u>FACTS TO KNOW BEFORE FINDING ANSWER :-</u>

  • Chemical formula of Propan-1-ol = CH₃CH₂CH₂OH [or] C₃H₈O
  • Molar mass of C = 12g/mol.
  • Molar mass of H = 1g/mol.
  • Molar mass of O = 16g/mol.

<u>SOLUTION :-</u>

Molar mass of C₃H₈O = 3(Molar mass of C) + 8(Molar mass of H) + 1(Molar mass of O)

⇒ Molar mass of C₃H₈O = (3×12) + (8×1) + 16 = 36 + 8 + 16 = <u>60g/mol.</u>

Ira Lisetskai [31]3 years ago
4 0

Answer:

60.0952 g/mol

Formula: C₃H₈O

Density: 803 kg/m³

Melting point: -126 °C

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Serhud [2]

^{85}_{37}Rb has 37 electrons in the orbit about its nucleus.

<h3>How to write an atomic symbol with atomic mass and atomic number?</h3>

Atomic mass is the total number of neutrons and protons present in the nucleus of an element.

The atomic number is the total number of protons present in the nucleus of an element.

For writing an atomic symbol, atomic mass is written on the top left of the atomic symbol, and the atomic number is written on the downside of the left of the symbol.

^{Atomic mass}_{Atomic number}X

An electrically neutral atom is an atom that has an equal number of electrons as well as protons.

Thus from the above conclusion we can say that  ^{85}_{37}Rb has 37 electrons.

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3 0
2 years ago
How many of the following are found in 15.0 kmol of xylene (C8H10)? (a) kg C8H10; (b) mol C8H10; (c) lb-mole C8H10; (d) mol (g-a
77julia77 [94]

Answer:

a) 1592.4 kg C8H10

b) 15*10³ mol C8H10

c) 33.1 lb-mole

d) 1.2 *10^5 mol C

e)  1.5 * 10^5 mol H

f)  1.44 * 10^6 grams C

g) 1.52 * 10^5 grams H

h) 9.03*10^27 molecules C8H10

Explanation:

Step 1: Data given

Number of moles C8H10 = 15.0 kmol =15000 moles

Molar mass of C8H10 = 106.16 g/mol

Step 2: Calculate mass C8H10

Mass C8H10 = moles C8H10 * molar mass C8H10

Mass C8H10 = 15000 * 106.16 g/mol

Mass C8H10 = 1592400 grams = 1592.4 kg

Step 3: Calculate moles C8H10

15.0 kmol = 15*10³ mol C8H10

Step 4: Calculate lb-mol

15.0 kmol = 33.1 lb-mole

Step 5: Calculate moles of C

For 1 mol C8H10 we have 8 moles of C

For 15*10³ mol C8H10 we have 8* 15*10³  =1.2 *10^5 mol C

Step 6: Calculate moles H

For 1 mol C8H10 we have 10 moles of H

For 15*10³ mol C8H10 we have 10* 15*10³  = 1.5 * 10^5 mol H

Step 7: Calculate mass of C

Mass C = moles C * molar mass C

Mass C = 1.2 *10^5 mol C * 12 g/mol

Mass C = 1.44 * 10^6 grams

Step 8: Calculate mass of H

Mass H = moles H * molar mass H

Mass H = 1.5 *10^5 mol H * 1.01 g/mol

Mass C = 1.52 * 10^5 grams

Step 9: Calculate molecules of C8H10

Number of molecules = number of moles * number of Avogadro

Number of molecules =  15*10³ mol C8H10 * 6.022*10^23

Number of molecules = 9.03*10^27 molecules

8 0
3 years ago
Be<br><br> Name the element.<br><br> Number of shells?<br><br> Valence electrons?
Vedmedyk [2.9K]

Answer:

Name the element: Beryllium

Number of shells:  4

Valence electrons: 2

Explanation:

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What a three of the signs of an allergic reaction to iv t-pa?
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The mass unit associated with density is usually grams. if the volume (in ml or cm3) is multiplied by the density (g/ml or g/cm3
jok3333 [9.3K]

The weight in grams = 7.93 g

Given volume = 2.00 in^{3}

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We need to find the Mass(weight) in grams.

To find the weight in grams we need to keep in mind that the volume and density must use the same volume unit for cancellation. So that the volume units will cancel out, leaving only the mass units.

The unit of given volume is in^{3} and unit of volume in density is cm^{3} , so first we need to change the unit of volume from in^{3} to cm^{3} so that the volume units will cancel out, leaving only the mass units.

1 in^{3} = 16.39 cm^{3} (given conversion)

2 in^{3}\times \frac{(16.39 cm^{3})}{(1 in^{3})}

in^{3} units get cancel out leaving the cm^{3} unit.

2 in^{3} = 32.78 cm^{3}

Mass = Density X Volume.

Density = 0.242 g/cm^{3} and Volume = 32.78 cm^{3}

Mass =0.242\frac{g}{cm^{3}}\times 32.78 cm^{3}

Mass = 7.93 grams (g)


3 0
3 years ago
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