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serg [7]
3 years ago
5

Which of the following processes will determine the number of moles in a sample? Dividing the mass of the sample by Avogadro's n

umber Multiplying the mass of the sample by Avogadro's number Dividing the number of molecules in the sample by Avogadro's number Multiplying the number of molecules in the sample by Avogadro's number
Chemistry
1 answer:
Olenka [21]3 years ago
6 0

The answer is: Dividing the number of molecules in the sample by Avogadro's number.

The Avogadro’s number is the number of atoms in 12 grams of the isotope carbon-12 (¹²C).

Na is Avogadro number or Avogadro constant (the number of particles, in this example carbon, that are contained in the amount of substance given by one mole).

The Avogadro number has value 6.022·10²³ 1/mol in the International System of Units; Na = 6.022·10²³ 1/mol.

For example:

N(Ba) = 2.62·10²³; number of atoms of barium.

n(Ba) = N(Ba) ÷ Na.

n(Ba) = 1.3·10²⁴ ÷ 6.022·10²³ 1/mol.

n(Ba) = 2.158 mol; amount of substance of barium.

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2 years ago
Red #40 has an acute oral LD50 of roughly 5000 mg dye/1 kg body weight. This means if you had a mass of 1 kg, ingesting 5000 mg
FrozenT [24]

Answer:

350 g dye

0.705 mol

2.9 × 10⁴ L

Explanation:

The lethal dose 50 (LD50) for the dye is 5000 mg dye/ 1 kg body weight. The amount of dye that would be needed to reach the LD50 of a 70 kg person is:

70 kg body weight × (5000 mg dye/ 1 kg body weight) = 3.5 × 10⁵ mg dye = 350 g dye

The molar mass of the dye is 496.42 g/mol. The moles represented by 350 g are:

350 g × (1 mol / 496.42 g) = 0.705 mol

The concentration of Red #40 dye in a sports drink is around 12 mg/L. The volume of drink required to achieve this mass of the dye is:

3.5 × 10⁵ mg × (1 L / 12 mg) = 2.9 × 10⁴ L

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3 years ago
To prevent poisoning, what three items should NEVER be taken into the lab?
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Read 2 more answers
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
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