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2 years ago

Convert 19 kcal into joules

Chemistry

1 answer:

il63 [147K]2 years ago

7 0

$19~ \text{kcal} = 19 \times 1000 ~\text{cal } = 19000~ \text{cal} = 19000 \times 4.184~ \text{J} = 79496~ \text{J}$

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A 265-mL flask contains pure helium at a pressure of 751 torrs. A second flask with a volume of 465 mL contains pure argon at a

Nadya [2.5K]

Answer:

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar = 472.8 torr

Explanation:

Step 1: Data given

Volume of the flask helium = 265 mL

Pressure in the helium flask = 751 torr = 751/760 atm

Volume of the flask argon = 465 mL

Pressure in the argon flask = 727 torr = 727/760 atm

The total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component in a gas mixture.

Step 2: Calculate total volume

Total volume = 265 mL + 465 mL = 730 mL = 0.730 L

Step 3: Boyle's Law:

P1V1=P2V2

⇒ with P1 = total pressure gas exerts in its own flask

⇒ with V1 = volume of flask with stopcock valve closed

⇒ with P2 = partial pressure of gas exerts on total volume of both flasks when stopcock valve is opened

⇒ with V2 = total volume of both flasks with stopcock valve opened

Helium using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of helium = 751 /760 = 0.98816 atm

⇒ with V1 = volume of helium = 0.265 L

⇒ with P2 = The new partial pressure of helium

⇒ with V2 = total volume = 0.730 L

(0.98816 atm)(0.265L)=P2(0.730L)

P2=0.359 atm

Argon using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of argon = 727/760 = 0.95658 atm

⇒ with V1 = volume of argon = 0.465 L

⇒ with P2 = The new partial pressure of argon

⇒ with V2 = total volume = 0.730 L

(0.95658 atm)(0.465L)=P2(0.730L)

P2=0.609 atm

Step 4: Convert pressure in atm to torr

Pressure helium = 0.359 atm = 272.8 torr

Pressure argon = 0.609 atm = 472.8 torr

Step 5: Calculate Total pressure

Ptotal = P(He)+P(Ar)

⇒ Pt = total pressure of the gas mixture

⇒ P(He) = partial pressure of Helium

⇒ P(Ar) = partial pressure of Argon

Pt = 272.8 torr + 472.8 torr

Pt = 745.6 torr

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar = 472.8 torr

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3 years ago

A sample of iron has the dimensions of 2 cm x 3 cm x 2 cm. If the mass of this

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Answer: Volume iron = 2 x 3 x 2 => 12 cm³

Mass = 94 g

D = m / V

D = 94 / 12

D = 7.833 g/cm³

Explanation:

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3 years ago

How many sigma bonds does CS2 have?

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Answer:

2 sigma bonds

CS2 is called carbon disulphide. The 2 unpaired electrons of each sulphur atom get paired with the 4 unpaired electrons of carbon. But it is a double bond. That is, 2 sigma bonds (sigma s-p , sigma p-p) and 2 pi (πp-p) bonds.

Explanation:

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3 years ago

Calcula el pH de estas soluciones: A-acido sulfúrico 0,0013 m B- hidróxido de socio 1,7 . 10 (-3) C- hidróxido de bario 2,2 . 10

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Answer:

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3 years ago

Palladium (Pd; Z 46) is diamagnetic. Draw partial orbital diagrams to show which of the following electron configurations is con

nexus9112 [7]

Answer : The electron configurations consistent with this fact is, (b) [Kr] 4d¹⁰

Explanation :

Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom are determined by the electronic configuration.

Paramagnetic compounds : They have unpaired electrons.

Diamagnetic compounds : They have no unpaired electrons that means all are paired.

The given electron configurations of Palladium are:

(a) [Kr] 5s²4d⁸

In this, there are 2 electrons in 's' orbital and 8 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital are paired but 'd' orbital are not paired. So, this configuration shows paramagnetic.

(b) [Kr] 4d¹⁰

In this, there are 10 electrons in 'd' orbital. From the partial orbital diagrams we conclude that electrons in 'd' orbital are paired. So, this configuration shows diamagnetic.

(c) [Kr] 5s¹4d⁹

In this, there are 1 electron in 's' orbital and 9 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital and 'd' orbital are not paired. So, this configuration shows paramagnetic.

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3 years ago