1. when energy increases, amplitude ________

The lab procedure you just studied involves

Svetllana [295]

Answer: c , v, c, c, v

Explanation:

8 0

2 years ago

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i

Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane ⇄ Isobutane

At equilibrium

1.0 M 2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M -

After equilibrium reestablishes:

(1.50-x)M (2.5+x)

The equilibrium expression will wriiten as:

$K_c=\frac{[Isobutane]}{[Butane]}$

$2.5=\frac{2.5+x}{(1.50-x)}$

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

3 0

2 years ago

Write a balanced net ionic equation for the following reaction: SrCl2(aq) + H2CO3(aq) → SrCO3(s) + HCl (aq)

max2010maxim [7]

We will balance the equation in the following order: metals, amethals, carbon, hydrogen and oxygen (the most common order).
The metal present in the equation is Sr, which is already balanced (there are 1 on each side of the equation).
The amethal present in the equation is Cl. There is 2 Cl in the left side and only one in the right side. So, we will multiply the quantity of the molecule that contains Cl by 2. Doing this, we'll obtain:

$SrCl_2_{(aq)}+H_2CO_3_{(aq)}\to SrCO_3_{(s)}+2HCl_{(aq)}$

Looking at the equation, we can see that it is now fully balanced. Hence, a balanced equation of the reaction is:

$SrCl_2_{(aq)}+H_2CO_3_{(aq)}\to SrCO_3_{(s)}+2HCl_{(aq)}$

5 0

2 years ago

How many molecules in each sample? 64.7 g N2 83 g CCl4 19 g C6H12O6

lilavasa [31]

Answer:

1.39x10²⁴ molecules N₂

.25x10²³ molecules CCl₄

6.38x10²² molecules C₆H₁₂O₆

Explanation:

First we convert the given masses into moles, using the compounds' respective molar mass:

64.7 g N₂ ÷ 28 g/mol = 2.31 mol N₂

83 g CCl₄ ÷ 153.82 g/mol = 0.540 mol CCl₄

19 g C₆H₁₂O₆ ÷ 180 g/mol = 0.106 mol C₆H₁₂O₆

Then we multiply each amount by Avogadro's number, to calculate the number of molecules:

2.31 mol N₂ * 6.023x10²³ molecules/mol = 1.39x10²⁴ molecules

0.540 mol CCl₄ * 6.023x10²³ molecules/mol = 3.25x10²³ molecules

0.106 mol C₆H₁₂O₆ * 6.023x10²³ molecules/mol = 6.38x10²² molecules

3 0

2 years ago

The density of gold is 19.3 g /cm cubed The density of iron pyrite is 5.0 g /cm cubed. Is a nugget of iron pyrite and a nugget o

telo118 [61]

If you clear volume in the density equation:

$\rho = \frac{m}{V}\ \to\ V = \frac{m}{\rho}$

The greater the density the lower the volume. This means, the volume of gold nugget will be smaller than the volume of iron pyrite nugget.

$V_{gold} = \frac{m}{\rho} = \frac{50\ g}{19.3\ g/cm^3} = \bf 2.59\ cm^3$

$V_{iron} = \frac{m}{\rho} = \frac{50\ g}{5.0\ g/cm^3} = \bf 10\ cm^3$

6 0

3 years ago

1. when energy increases, amplitude _________.