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2 years ago

A container holds 7.5 liters of co2 at STP, how many moles of gas is this ?

Chemistry

2 answers:

HACTEHA [7]2 years ago

8 0

Answer:- There are 0.337 moles of the gas.

Solution:- Volume(V) of carbon dioxide gas = 7.5 L

Conditions are STP means, T = 273 K and P = 1 atm

From ideal gas law equation, PV = nRT

n = PV/RT

Where n is the number of moles and R is the universal gas constant and it's value is 0.0821 atm.L/mol.K

Let's plug in the values in the formula..

n = (1 x 7.5)/(0.0821 x 271)

n = 0.337 mol

photoshop1234 [79]2 years ago

4 0

One mole of any ideal gas at STP conditions will have a volume equal to 22.4L

7.5 L of CO2 /22.4 L = 0.3348 moles = 0.33 mole

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natka813 [3]

930 g for 15 clementines

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2 years ago

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a gas sample is heated from -20.0 C to 57.0 C and the volume is increased from 2.00L to 4.50L. If the initial pressure is 0.109a

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2 years ago

What mass of powdered drink mix is needed to make a 0.5 M solution of 100 mL?

Maru [420]

Answer:

There is 17,114825 g of powdered drink mix needed

Explanation:

<u>Step 1 :</u> Calculate moles

As given, the concentration of the drink is 0.5 M, this means 0.5 mol / L

Since the volume is 100mL, we have to convert the concentration,

⇒0.5 / 1 = x /0.1 ⇒ 0.5* 0.1 = x = 0.05 M

This means there is 0.05 mol per 100mL

<u>Step 2 </u>: calculate mass of the powdered drink

here we use the formula n (mole) = m(mass) / M (Molar mass)

⇒ since powdered drink mix is usually made of sucrose (C12H22O11) and has a molar mass of 342.2965 g/mol.

0.05 mol = mass / 342.2965 g/mol

To find the mass, we isolate it ⇒0.05 mol * 342.2965 g/mol = 17,114825g

There is 17,114825 g of powdered drink mix needed

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2 years ago

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The metallic radius of a lithium atom is 152 pm. What is the volume of a lithium atom in cubic meters?

Anuta_ua [19.1K]

Answer:

Volume of lithium atom is found to be 1.47 X 10⁻²⁹ m³

Explanation:

Let us consider the volume of atom as a sphere (but it is little complex than that). This volume is mathematically expressed as,

$V=\frac{4}{3}\pi R^{3}$ ----------------------------------------------------------------------------------------(Eq. 1)

Here, R is the radius of lithium atom. The radius is given in picometers, so firstly let us convert it into meters

$R=(152pm )(1X10^{-12}\frac{m}{pm})$

$R = 1.52 X 10^{-10}m$

placing this value in Eq.1 the required result is achieved

$V=\frac{4}{3}\pi {1.52X10^{-10}}^{3}$

V= 1.47 X 10⁻²⁹ m³

8 0

2 years ago

Be sure to answer all parts. For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g)

Liono4ka [1.6K]

Answer:

a. -268.13 J/K

b. -279.95 J/K

c. + 972.59 J/K

Explanation:

The value of the change in entropy (ΔS°) can be calculated by:

ΔS° = ∑n*S° products - ∑n*S° reactants, where n is the stoichiometric number of moles.

The values of S° for each substance can be found on a thermodynamic table.

a. 3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)

S°, NO2(g) = 240.06 J/mol.K

S°, H2O(l) = 69.91 J/mol.K

S°, HNO3(l) = 155.60 J/mol.K

S°, NO(g) = 210.76 J/mol.K

ΔS° = (210.76 + 2*155.60) - (3*240.06 + 69.91)

ΔS° = -268.13 J/K

b. N2(g) + 3F2(g) → 2NF3(g)

S°, N2(g) = 191.61 J/mol.K

S°, F2(g) = 202.78 J/mol.K

S°, NF3(g) = 260.0 J/mol.K

ΔS° = (2*260.0 ) - (191.61 + 3*202.78)

ΔS° = -279.95 J/K

c. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g)

S°, C6H12O6(s) = 212 J/mol.K

S°, O2(g) = 205.138 J/mol.K

S°, CO2(g) = 213.74 J/mol.K

S°, H2O(g) = 188.83 J/mol.K

ΔS° = (6*213.74 + 6*188.83) - (212 + 6*205.138)

ΔS° = +972.59 J/K

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2 years ago