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3 years ago

What volume would 3.00 moles of neon gas have at 295 K and 645 mmHg?

Chemistry

1 answer:

LUCKY_DIMON [66]3 years ago

6 0

Answer:

V = 85.619 L

Explanation:

To solve, we can use the ideal gas law equation, PV = nRT.

P = pressure (645 mmHg)

V = volume (?)

n = amount of substance (3.00 mol)

R = ideal gas constant (62.4 L mmHg/mole K)

T = temperature (295K)

Now we would plug in the appropriate numbers into the equation using the information given and solve for V.

(645)(V) = (3.00)(62.4)(295)

(V) = (3.00)(62.4)(295)/645

V = 85.619 L

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3 years ago

Given the following, determine ΔG°f at 298 K for SnO. Sn(s) + SnO2(s) → 2SnO(s) ; ΔG° = 12.0 kJ at 298K

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Answer:

The value of change in Gibbs free energy for tin(II) oxide solid is -251.9 kJ/mol.

Explanation:

$Sn(s) + SnO_2(s)\rightarrow 2SnO(s), \Delta G_{f}^{o} = 12.0 kJ$

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$12.0 kJ=(2 mol\times \Delta G_{f,SnO}^{o})-(1mol\times 0 kJ/mol+1 mol\times -515.8 kJ/mol)$

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The value of change in Gibbs free energy for tin(II) oxide solid is -251.9 kJ/mol.

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3 years ago

2mL of a serum sample was added to 18mL of phosphate buffered saline (PBS) in Tube 1. 10mL of Tube 1 was added to 40mL of PBS in

Sladkaya [172]

Answer:

Tube 2 has a total dilution of 1:50

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We have a 2 ml serum sample added to a 18 mL phosphate buffered saline sample in tube 1. This means now in tube 1 there is 20 mL.

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10 ml of this 1:10 diluted tube 1 is taken and added to a 40 mL of PBS in tube 2.

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Tube 2 has a total dilution of 1:50

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3 years ago

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Nookie1986 [14]

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2 years ago

Read 2 more answers

g Sucrose (C12H22O11), a nonionic solute, dissolves in water (normal freezing/melting point 0.0°C) to form a solution. If some u

DIA [1.3K]

Answer:

15.4 g of sucrose

Explanation:

Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m

0.56°C / 1.86 m/°C = m → 0.301 mol/kg

m → molality (moles of solute in 1kg of solvent)

Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg

0.301 mol/kg . 0.150kg = 0.045 moles.

We determine the mass of sucrose, by the molar mass:

0.045 mol . 342 g/1mol = 15.4 g

4 0

3 years ago