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3 years ago

How many aluminium atoms are there in the formula for aluminium nitrate ?

1 answer:

8 0

The formula for Aluminium Nitrate is $Al(NO _{3} )_{3}$ .

This is because aluminium has a 3+ charge and nitrate has a 1- charge, so you need 3 of them to balance out aluminium.

Therefore, you have one ion of aluminium.

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A 0.15 mol sample of H2S is placed in a 10 L reaction vessel and heated to 1132◦C. At equilibrium, 0.03 mol H2is present. Calcul

Angelina_Jolie [31]

Answer: The value of the equilibrium constant is 0.00009.

Explanation:

Initial moles of $H_2S$ = 0.15 mole

Volume of container = 10 L

Initial concentration of $H_2S=\frac{moles}{volume}=\frac{0.15moles}{10L}=0.015M$

Moles of $H_2$ at equilibrium= 0.03 mole

equilibrium concentration of $H_2=\frac{moles}{volume}=\frac{0.03moles}{10L}=0.003M$ [/tex]

The given balanced equilibrium reaction is,

$2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)$

Initial conc. 0.015 M 0 0

At eqm. conc. (0.015-2x) M (2x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}$

$K_c=\frac{(2x)^2\times x}{(0.015-2x)^2}$

we are given : 2x= 0.003 M

x= 0.0015 M

Now put all the given values in this expression, we get :

$K_c=\frac{(0.003)^2\times 0.0015}{(0.015-0.003)^2}$

$K_c=0.00009$

Thus the value of the equilibrium constant is 0.00009.

8 0

3 years ago

A solution contains 482.6 g

mariarad [96]

Answer:

$\large \boxed{\text{102.57 $\, ^{\circ}$C}}$

Explanation:

Data:

m₂ = 482.6 g sucrose

m₁ = 0.2804 kg water

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{\text{b}} = iK_{\text{b}}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For sucrose, i = 1.

1. Moles of sucrose

$n = \text{482.6 g} \times \dfrac{\text{1 mol}}{\text{342.30 g}} = \text{1.410 mol}$

2. Molal concentration of sucrose

$b = \dfrac{\text{1.410 mol}}{\text{0.2804 kg}} = \text{5.029 mol/kg}$

3. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 5.029 \cdot mol \cdot kg^{-1} = 2.574 \, ^{\circ}\text{C}$

4. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 2.574 \, ^{\circ}\text{C} = \mathbf{102.57 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{102.57 \, ^{\circ}C}}$}$

7 0

3 years ago

What happen to the wind turbines to cause some people to lose electricity?

madreJ [45]

They may be frozen over

6 0

3 years ago

In the gas phase, the energy difference between a cesium atom and its most stable ion is referred to as

Sholpan [36]

Cesium is in the first group, so it forms an ion with a charge of +1. The change that cesium atoms undergo when they are in the gaseous state to form gaseous ions is known as the ionization energy. The ionization energy, by definition, is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of +1 ions from the atoms.

6 0

4 years ago

Hey! Please help. Will mark brainlist if u are right, ty

Tema [17]

Answer:

40 hopefully right

Explanation:

8 0

3 years ago