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2 years ago

In a fixed cylinder are 3moles of oxygen gas at 300Kelvin and 1.25atm. What is the volume of the container?

Chemistry

1 answer:

vladimir2022 [97]2 years ago

6 0

Answer:

The volume of the container is 59.112 L

Explanation:

Given that,

Number of moles of Oxygen, n = 3

Temperature of the gas, T = 300 K

Pressure of the gas, P = 1.25 atm

We need to find the volume of the container. For a gas, we know that,

PV = nRT

V is volume

R is gas constant, R = 0.0821 atm-L/mol-K

So,

$V=\dfrac{nRT}{P}\\\\V=\dfrac{3\ mol\times 0.0821\ L-atm/mol-K \times 300\ K}{1.25\ atm}\\\\V=59.112\ L$

So, the volume of the container is 59.112 L

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Give the coordination number, the charge of the central metal ion, and select the correct name in each coordination compound: A.

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The correct question is

Give the coordination number, the charge of the central metal ion, and select the correct name in each coordination compound:

Na3[CoCl6]

[Ni(CO)4]

[Ni(NH3)3(H2O)2] (NO3)2

Answer:

See explanation for details

Explanation:

Na3[CoCl6]

Name: Sodium hexacholorocobalt III

Charge on the complex: -3

Central metal ion name : cobalt III

coordination number: 6

[Ni(CO)4]

Name: tetracarbonyl nickel (0)

Charge on the complex: 0

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coordination number: 4

[Ni(NH3)3(H2O)2] (NO3)2

Name: Diaquatriaminenickel II nitrate

Charge on the complex: +2

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4 0

2 years ago

Which compound would be expected to show intense IR absorption at 2710 and 1705 cm-1? (Ph = phenyl group)

Serhud [2]

Answer:

B. PhCHO

Explanation:

Every organic group shows a characteristic IR absorption at certain wavelength . With the help of these absorption spectra we can identify the group present on organic molecules .

The wave number of 2710 cm⁻¹ is absorbed by aldehyde bond stretching .

The wave number of 1705 cm⁻¹ is shown by conjugated aldehyde . So the most likely compound among given compounds is PhCHO .

7 0

2 years ago

Consider the dissolution of 1.50 grams of salt XY in 75.0 mL of water within a calorimeter. The temperature of the water decreas

-BARSIC- [3]

Answer:

The quantity of heat lost by the surroundings is 258,5J

Explanation:

The dissolution of salt XY is endothermic because the water temperature decreased.

The total heat consumed by the dissolution process is:

4,184 J/g°C × (75,0 + 1,50 g) × 0,93°C = 297,7 J

This heat is consumed by the calorimeter and by the surroundings.

The heat consumed by the calorimeter is:

42,2 J/°C × (0,93°C) = 39,2 J

That means that the quantity of heat lost by the surroundings is:

297,7J - 39,2J = 258,5 J

I hope it helps!

8 0

2 years ago

According to the equation below, how many moles of Ca(OH)2 are required to react with 1.36 mol H3PO4 to produce Ca3(PO4)2? 3Ca(O

allsm [11]

Answer: The amount of calcium hydroxide needed to react is 2.04 moles

Explanation:

We are given:

Moles of phosphoric acid = 1.36 moles

For the given chemical equation:

$3Ca(OH)_2+2H_3PO_4\rightarrow Ca_3(PO_4)_2+6H_2O$

By Stoichiometry of the reaction:

2 moles of phosphoric acid reacts with 3 moles of calcium hydroxide

So, 1.36 moles of phosphoric acid will react with = $\frac{3}{2}\times 1.36=2.04mol$ of calcium hydroxide

Hence, the amount of calcium hydroxide needed to react is 2.04 moles

3 0

2 years ago

495 cm3 of oxygen gas and 877 cm3 of nitrogen gas, both at 25.0 C and 114.7 kpa, are injected into an evacuated 536 cm3 flask. F

I am Lyosha [343]

Answer:

Total pressure of the flask is 2.8999 atm.

Explanation:

Given data:

Volume of oxygen (O2) gas= 495 cm3

= 0.495 L (1 cm³ = 1 mL = 0.001 L)

Volume of nitrogen (N2) gas = 877 cm3

= 0.877 L (1 cm³ = 1 mL = 0.001 L)

volume of falsk = 536 cm3

= 0.536 L (1 cm³ = 1 mL = 0.001 L)

Temperature = 25 °C

T = (25°C + 273.15) K

= 298.15 K

Pressure = 114.7 kPa

= 114.700 Pa

Pressure (torr) = 114,700 / 101325

= 1.132 atm

Formula:

PV=nRT (ideal gas equation)

P = pressure

V = volume

R (gas constnt)= 0.0821 L.atm/K.mol

T = temperature

n = number of moles for both gases

Solution:

Firstly we will find the number of moles for oxygen and nitrogen gas.

For Oxygen:

n = PV / RT

n = 1.132 atm × 0.495 L / 0.0821 L.atm/K.mol × 298.15 K

= 0.560 / 24.47

= 0.0229 moles

For Nitrogen:

n = PV / RT

n = 1.132 atm × 0.877 / 0.0821 L.atm/K.mol × 298.15 K

n = 0.992 / 24.47

= 0.0406

Total moles = moles for oxygen gas + moles for nitrogen gas

= 0.0229 moles + 0.0406 moles

n = 0.0635 moles

Now put the values in formula

PV=nRT

P = nRT / V

P = 0.0635 × 0.0821 L.atm/K.mol × 298.15 K / 0.536 L

P = 1.554 / 0.536

P = 2.8999 atm

Total pressure in the flask is 2.8999 atm, while assuming the temperature constant.

7 0

2 years ago

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