Lithium and oxygen only exist in nature as Diatomic Molecules. True or False

2.3 x 10²¹ particles H = __________ mol H (round to the nearest thousandth)

Natalka [10]

Taking into account the definition of avogadro's number, 3.82×10⁻³ moles of H are 2.3×10²¹ particles of H.

<h3>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.

Then you can apply the following rule of three: if 6.023×10²³ particles are contained in 1 mole of H, then 2.3×10²¹ particles are contained in how many moles of H?

amount of moles of H= (2.3×10²¹ particles × 1 mole)÷ 6.023×10²³ particles

<u><em>amount of moles of H= 3.82×10⁻³ moles</em></u>

Finally, 3.82×10⁻³ moles of H are 2.3×10²¹ particles of H.

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7 0

2 years ago

Help with this question

NeTakaya

Isotope atoms mean, atoms with the same number of protons but different number of neutrons in their nuclei . The given oxygen atoms are isotopes meaning they have different number of neutrons in their nuclei . Oxygen 16 have 8 protons and 8 neutrons in its nucleus but Oxygen has 8 protons and 10 neutrons. That's what make them different.

3 0

2 years ago

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2.6(b) A sample of 2.00 mol CH3OH(g) is condensed isothermally and reversibly to liquid at 64°C. The standard enthalpy of vapori

Sophie [7]

Answer:

The value of W is 5.602 kJ, Q is -70.6 kJ, change in U is -65 kJ, and change in H is -70.3 kJ.

Explanation:

Based on the given information, the mass of CH3OH given is 64 grams, which is condensed isothermally and reversibly to liquid at 64 degrees C. The given standard enthalpy of vaporization of methanol at 64 degrees C is 35.3 kJ per mole.

The moles of CH3OH can be determined by using the formula,

Moles = Mass / Molar mass

= 64.0 grams / 32.0 grams per mole

= 2 mol

The amount of energy given by the process of condensation is,

ΔH = 2 mol × 35.3 kJ/mol = 70.6 kJ

In condensation heat is given off, thus, it is an exothermic process, hence, q will be -70.6 kJ

The work or W can be calculated by using the formula,

W = -P ΔV

Let us first find the volume of 2.0 mole gas at 64 °C, or 64 + 273 = 337 K,

PV = nRT

V = nRT/P

= 2 mol × 0.08206 L atm per mol K × 337 K/1 atm

= 55.3 L

As the liquid condenses in the process, the change in volume would be negligible. So, the volume change will be -55.3 L

W = - 1 atm × - 55.3 L

W = 55.3 L.atm

W = 55.3 L.atm × 101.3 J/1 L atm = 5602 J

W = 5602 × 1 kJ / 1000 J = 5.602 kJ

W = 5.602 kJ

Now U can be calculated using the formula,

U = q + W

= -70.6 kJ + 5.602 kJ

= -65. kJ

Thus, q = -70.6 kJ, W = 5.602 kJ, U = -65 kJ, and ΔH = -70.3 kJ.

4 0

3 years ago

Assume that 1.0 mol of C4H10 is completely burned in excess oxygen to form carbon dioxide and water. How many moles of CO2 would

In-s [12.5K]

You need to first write a chemical equation and balance it
C₄H₁₀ + O₂ → CO₂ + H₂O
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
1.0 moles X moles
1.0 mol C₄H₁₀ ( $\frac{8 mol CO₂}{2 mol C₄H₁₀}$ ) = 4 moles of CO₂

7 0

2 years ago

In the peptide GLVIW, the C-terminal end is ________.

zheka24 [161]

The peptide given above is made up of the following amino acids: glycine [G], leucine [L], valine [V], isoleucine [I] and tryptophan [W]. These amino acids are joined together by amide bond to form peptide. Peptides usually have two terminals, the N terminal and the C terminal. For GLVIW, the C terminal end amino acid is tryptophan, that is the last amino acid on the peptide chain. The N terminal amino acid is glycine, that is, the first amino acid on the peptide chain.

5 0

2 years ago