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3 years ago

Which is the molar mass of BF3?

Chemistry

2 answers:

san4es73 [151]3 years ago

7 0

Boron trifluoride 68.005

anygoal [31]3 years ago

3 0

Boron trifluoride/Molar mass
67.81 g/mol

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Describe the difference between cations represented on group 1/2(red and orange)and those found on the transition metal group(pe

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3 years ago

Name the three types of scientific investigations, and briefly describe each.

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Descriptive, Comparative, and experimental.

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4 years ago

At a certain temperature the vapor pressure of pure heptane

Amanda [17]

Answer:

215 mm Hg

Explanation:

This is a problem we can solve by utilizing Raoult´s law which states that the partial pressure of component A, P (A) in a solution is given by:

P (A) = X (A ) Pº(A)

where A is the mol fraction of component A, and Pº (A) is the pressure of pure component A.

Now here we have the masses so we can calculate the number of moles and the mol fraction of component A

n = number of moles = mass / MW

n( C₇H₁₆ ) = 102 g / 100.21 g/mol = 1.02 mol

n( CHCl₃ ) = 135 g / 119.38 g/mol ) = 1.13 mol

The mol fraction of C₇H₁₆ is given by:

X (C₇H₁₆) = n( C₇H₁₆ )/ [n( C₇H₁₆ ) + n( CHCl₃ )]

= 1.02 / ( 1.02 + 1.13 ) = 0.474

Then

P(C₇H₁₆) = X (C₇H₁₆) x (C₇H₁₆)

= 0.474 x 454 mm Hg = 215 mm Hg

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3 years ago

A piston-cylinder device has two stops; a lower set and an upper set, that constrain the cylinder. When the piston is on the low

stich3 [128]

Answer:

2080kJ

Explanation:

m(U4 - U1) = Q4 - W4

If P is less than 300kpa, then V = 0.4L

If P is greater than 300kpa, then V = 0.6L

If P is equal to 300kpa, then 0.4L < V < 0.6L

At 100kpa, vf = 0.001043m/kg and vg = 1.693m³/kg

Level 1

v1 = 0.001043 + 0.2 x 1.693 = 0.33964

m = V1/v1 ⇔ 0.4/0.33964

m = 1.178kg

u1 = 417.36 + 0.2 x 2088.7 = 835.1kJ/kg

Level 3

v3 = 0.6/1.178 = 0.5095 < vG = 0.6058 at P3 is equal to 300kpa

NOTE: Piston does reach upper stops to reach saturated vapor

Level 4

v4 = v3 = 0.5095m³/kg = vG at P4

P4 = 361kpa, u4 = 2550.0kJ/kg

1W4 = 1W2 + 2W3 + 3W4 = 0 + 2W3 + 0

1W4 = P2 (V3 - V2) = 300 x (0.6 - 0.4) = 60kJ

1Q4 = m(u4 -u1) + 1W4 = 1.178(2550.0 - 835.1) + 60 = 2080kJ

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3 years ago