What type of wave is shown above transverse wave

Calculate the molar concentrations of H+ and OH− in solutions that have the following pH values:

Anastaziya [24]

Answer:

Explanation:

a )

pH = - log[ H⁺]

8.26 = - log[ H⁺]

[ H⁺] = 10⁻⁸°²⁶ mole / l

= 5.49 x 10⁻⁹ moles / l

[ H⁺] [OH⁻] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 5.49 x 10⁻⁹

= .182 x 10⁻⁵ moles / l

b )

10.25 = - log[ H⁺]

[ H⁺] = 10⁻¹⁰°²⁵ mole / l

= 5.62 x 10⁻¹¹ moles / l

[ H⁺] [OH⁻] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 5.62 x 10⁻¹¹

= .178 x 10⁻³ moles / l

c )

4.65 = - log[ H⁺]

[ H⁺] = 10⁻⁴°⁶⁵ mole / l

= 2.24 x 10⁻⁵ moles / l

[ H⁺] [OH⁻] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 2.24 x 10⁻⁵

= .4464 x 10⁻⁹ moles / l

7 0

3 years ago

The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular temperature: Time (s) (

harina [27]

Answer:

Part a: The rate of the equation for 1st order reaction is given as $Rate=k[H_2O_2]$

Part b: The integrated Rate Law is given as $[H_2O_2]=[H_2O_2]_0 e^{-kt}$

Part c: The value of rate constant is $7.8592 \times 10^{-4} s^{-1}$

Part d: Concentration after 4000 s is 0.043 M.

Explanation:

By plotting the relation between the natural log of concentration of $H_2O_2$ , the graph forms a straight line as indicated in the figure attached. This indicates that the reaction is of 1st order.

Part a

Rate Law

The rate of the equation for 1st order reaction is given as

$Rate=k[H_2O_2]$

Part b

Integrated Rate Law

The integrated Rate Law is given as

$[H_2O_2]=[H_2O_2]_0 e^{-kt}$

Part c

Value of the Rate Constant

Value of the rate constant is given by using the relation between 1st two observations i.e.

t1=0, M1=1.00

t2=120 s , M2=0.91

So k is calculated as

$-k(t_2-t_1)=ln{\frac{M_2}{M_1}}\\-k(120-0)=ln{\frac{0.91}{1.00}}\\k=\frac{-0.09431}{-120}\\k=7.8592 \times 10^{-4} s^{-1}$

The value of rate constant is $7.8592 \times 10^{-4} s^{-1}$

Part d

Concentration after 4000 s is given as

$-k(t_2-t_1)=ln{\frac{M_2}{1.0}}\\-7.8592 \times 10^{-4}(4000-0)=ln{\frac{M_2}{1.00}}\\-3.1437=ln{\frac{M_2}{1.00}}\\M_2=e^{-3.1437}\\M_2=0.043 M$

Concentration after 4000 s is 0.043 M.

7 0

3 years ago

For the reaction;

ankoles [38]

Answer:

At equilibrium:

[H2] = 0.005 M

[Br2] = 0.105 M

[HBr] = 0.189 M

Explanation:

H2(g) + Br2(g) ⇄ 2HBr

an "x" value will be used from reactant to produced "2x"

so at equilibrium:

[H2] = 0.1 - x

[Br2] = 0.2 - x

[HBr] = 2x

we know that Kc=[HBr]²/[H2][Br2]

Thus 62.5 = (2x)²/(0.1-x)(0.2-x)

this generate a quadratic equation: 58.5x² - 18.75x + 1.25 = 0

the x₁ = 0.23 x₂ = 0.09457

we pick 0.09457 because the two reactants can not make more than what they have. x₁ is higher than both initial reactant concentration

Then we substitute the "x₂" value at equilibrium:

[H2] = 0.1-0.09457 = 0.005 M

[Br2] = 0.2-0.09457 = 0.105 M

[HBr] = 2*0.09457 = 0.189 M

7 0

4 years ago

What element is a gas at room temp, has 6 valence electrons, and is non-reactive

sveta [45]

Answer:

Noble gases

Explanation:

Noble gases are nonreactive, nonmetallic elements in group 18 of the periodic table. As you can see in the periodic table in the figure below, noble gases include helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn).18 Jun 2021

7 0

3 years ago

Write some example of misicible and immisicible liquid

Eduardwww [97]

Explanation:

miscible liquids: water and alcohol

immiscible liquids: water and oil

4 0

3 years ago