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3 years ago

How are pressure and temperature related in terms of kinetic energy?

Chemistry

1 answer:

mylen [45]3 years ago

3 0

Answer:

As the temperature of the gas increases, the particles gain kinetic energy and their speed increases. This means that the particles hit off the sides more often and with greater force. Both of these factors cause the pressure of the gas to increase.

Explanation:

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The powder mixture (Cu, Al and Fe) = 10g was oxidized from sufficient chloride acid. 1) What are the possible reactions to the p

Mandarinka [93]

Answer:

1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

Explanation:

1) Possible reactions

2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Mass of each metal

a) Mass of Cu

The waste was the unreacted copper.

Mass of Cu = 2.5 g

b) Masses of Al and Fe

We have two relations :

Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g

H₂ from Al + H₂ from Fe = 6.38 L at NTP

i) Calculate the moles of H₂

NTP is 20 °C and 1 atm.

$\begin{array}{rcl}pV & = & n RT\\\text{1 atm} \times \text{6.38 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{293.15 K}\\6.38 & = & 24.06n \text{ mol}^{-1} \\n & = & \dfrac{6.38}{24.06 \text{ mol}^{-1} }\\\\ & = & \text{0.2652 mol}\\\end{array}$

(ii) Solve the relationship

Let x = mass of Al. Then

7.5 - x = mass of Fe

Moles of Al = x/27

Moles of Fe = (7.5 - x)/56

Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x /18

Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56

∴ x/18 + (7.5 - x)/56 = 0.2652

56x + 18(7.5 - x) = 267.3

56x + 135 - 18x = 267.3

38x = 132.3

x = 3.5 g

Mass of Al = 3.5 g

Mass of Fe = 7.5 g - 3.5 g = 4.0 g

The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

4 0

3 years ago

which of the following compounds exhibits dipole -dipole forces as its strongest attraction between molecules? o2,ch3Br,CCl4,He,

Archy [21]

Answer:

B. CH3Br

Explanation:

Dipole -Dipole interactions take place in polar molecules.

CH3Br exhibits dipole -dipole forces as its strongest attraction between molecules because it is a polar molecule due to the slightly negative dipole present on the Br molecule.

While O2 is a nonpolar molecule due to its linear structure, CCl4 has zero resultant dipole moment, Helium is non-polar and BrCH2CH2OH is a non polar compound having net dipole moment is zero.

Hence, the correct option is B. CH3Br.

3 0

3 years ago

La densidad del óxido de magnesio. MgO, es de 3.581 g/cm3 El MgO, es de 3.581 g/cm3 El MgO cristaliza con ordenamiento cúbico co

Jobisdone [24]

Answer:

$a=4.213cm$

$r=1.490x10^{-8}cm$

Explanation:

Hola.

En este caso, para calcular la longitud (a) de una cara de celda unitaria, consideramos la siguiente ecuación:

$\rho =\frac{#at*M}{a^3N_A}$

En la que consideramos el número de átomos por celda (4 para FCC), la masa molar (40.3 g/mol para MgO) y el número de avogadro para obtener:

$3.581g/mol = \frac{4atom/celda*40.3g/mol}{a^3*6.02x10^{23}atom/mol}$

Despejando para a, obtenemos:

$a^3 = \frac{4atom/celda*40.3g/mol}{3.581g/cm^3*6.02x10^{23}atom/mol}\\\\a=\sqrt[3]{7.478cm^3} \\\\a=4.213cm$

Finalmente, el radio lo calculamos como:

$r=\frac{\sqrt{2}*a}{4}=\frac{\sqrt{2}*4.213x10^{-8}cm}{4}\\\\r=1.490x10^{-8}cm$

¡Saludos!

6 0

3 years ago

For the following reaction, 33.7 grams of bromine are allowed to react with 13.0 grams of chlorine gas.

Ira Lisetskai [31]

Explanation:

The reaction is given as;

Br2(g) + Cl2(g) ----> 2BrCl(g)

From the equation;

1 mol of Br2 reacts with 1 mol of Cl2

Converting the masses given to moles, using the formular;

Number of moles = Mass / Molar mass

Br2;

Number of moles = 33.7 g / 159.808 g/mol = 0.21088 mol

Cl2;

Number of moles = 13.0 g / 70.906 g/mol = 0.18334 mol

From the values;

0.18334 mol of Cl2 would react with 0.18334 mol of Br2 with an excess of 0.02754 mol of Br2

<em>What is the maximum amount of bromine monochloride that can be formed? __________grams</em>

<em />

1 mol of Cl2 produces 2 mol of Bromine Monochloride

0.18334 mol of Cl2 would produce x

Solving for x;

x = 0.18334 * 2 = 0.36668 mol

Converting to mass;

Mass = Number of moles * Molar mass = 0.36668 mol * 115.357 g/mol

Mass = 42.299 g

<em />

<em>What is the FORMULA for the limiting reagent?</em>

<em />

The limiting reagent is Cl2 as it determines the amount of product formed. The moment the reaction uses up Cl2, the reaction stops.

<em>What amount of the excess reagent remains after the reaction is complete? __________grams</em>

The excess reagent is Br2

The number of moles left is;

0.02754 mol of Br2

Converting to mass;

Mass = Number of moles * Molar mass = 0.02754 mol * 159.808 g/mol

Mass = 4.401 g

7 0

2 years ago

1. What is the oxidation number of sulphur in Na2S2O8<br> A-9<br> B. -7<br> C. +7<br> D. +14

butalik [34]

Answer:

<h2>Hereis the correct answer </h2>

(A.-9)

Explanation:

<h3>STUDY CORRECTION. </h3>

6 0

2 years ago