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Klio2033 [76]
3 years ago
14

The element europium exists in nature as two isotopes: has a mass of u and has a mass of u. The average atomic mass of europium

is u. Calculate the relative abundance of the two europium isotopes.
Chemistry
1 answer:
ElenaW [278]3 years ago
4 0

Answer:

Problem Details

The element europium exists in nature as two isotopes:  151Eu has a mass of 150.9196 amu, and 153Eu has a mass of 152.9209 amu. The average atomic mass of europium is 151.96 amu. Calculate the relative abundance of the two europium isotopes.

answer:

151Eu = 48%, 153Eu = 52%

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nexus9112 [7]

Answer:

Cobalt chloride= D cobalt(111) chloride

5 0
4 years ago
Read 2 more answers
John dissolves .5g of a white powder in 25g of benzene (FP 5oC) (kf benzene is 5.1) and finds the solution freezes at 3.7oC. Det
navik [9.2K]

Answer:

The compound has a molar mass of 78.4 g/mol

Explanation:

Step 1: data given

Mass of a sample = 0.5 grams

Mass of benzene = 25 grams

Freezing poing = 5 °C

Kf of benzene = 5.1 °C/m

Freezing point solution = 3.7 °C

Step 2: Calculate molality

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 5.0 - 3.7 = 1.3 °C

⇒with i = the can't hoff factor = 1

⇒with Kf = the freezing point depression constant of benzene = 5.1 °C/m

⇒with m = the molality

1.3 = 5.1 * m

m = 1.3 / 5.1

m = 0.255 moles /kg

Step 3: Calculate moles

Molality = moles / mass benzene

0.255 molal = moles / 0.025 kg

Moles = 0.255 molal * 0.025 kg

Moles = 0.006375 moles

Step 4: Calculate molar mass of the compound

Molar mass compund = mass / moles

Molar mass compound = 0.5 grams / 0.006375 moles

Molar mass compound = 78.4 g/mol

The compound has a molar mass of 78.4 g/mol

7 0
3 years ago
What volume does 1.70 ×10–3 mol of chlorine gas occupy if its temperature is 20.2 °C and its pressure is 795 mm Hg?
Cerrena [4.2K]

Explanation:

The given data is as follows.

     No. of moles = 1.70 \times 10^{-3},           V = ?

     T = 20.2 + 273 K = 293.2 K,             P = \frac{795 mm Hg}{760.0 mm Hg/atm} = 1.046 atm,                      R = 0.0821 L atm K^{-1}mol ^{-1}

Calculate the volume using ideal gas equation as follows.

                                    P V = n R T

    1.046 atm \times V = 1.70 \times 10^{-3} \times 0.0821 L atm K^{-1}mol ^{-1} \times 293.2 K

                                  V = \frac{1.70 \times 10^{-3} \times 0.0821 L atm K^{-1}mol ^{-1} \times 293.2 K}{1.046 atm}

                                   = \frac{40.921 L atm}{1.046 atm}

                                   = 39.122 \times 10^{-3} L

Thus, we can conclude that volume of the gas is 39.122 \times 10^{-3} L.


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