The rate law for the reaction

How many moles are in 20 grams of argon

Sladkaya [172]

Answer:

There are 0.5 mole in 20g of argon.

Explanation:

40 g of argon = 1mole

Then 20g of argon is,

→ 1/40 × 20

→ 0.5 mole

5 0

3 years ago

HELP MEEEE ASAP!!! Answer these questions based on what you learned from the passage. You can assume any body structure comes fr

erica [24]

Answer:

1.Legs

2.Arms and hands

3.Armor

4.Skeletal

5.Carpenters

6.Brain

7.Skin

8.Blood

Explanation:

You can google dem and it will show you all the answers

4 0

4 years ago

Write the word and balanced chemical equations for the reaction between:

kupik [55]

Answer:

Alumminum hydroxide reacts with sulfuric acid as follows: 2Al(OH)3+H2SO4-->Al2(SO4)+6H2O.

Explanation:

plz mark as brainlist

7 0

3 years ago

Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:

Setler [38]

The order of frequency is d < a < c < b

$E_{n} = 13.6 * z^{2} / n^{2}$ eV

where z = atomic mass number

n = energy level

For hydrogen z = 1

Therefore, Energy for n = 1

$E_{n} = 13.6 * 1^{2} / 1^{2}$ eV

= -13.6 eV

for n = 2

$E_{n} = 13.6 * 1^{2} / 2^{2}$ eV

= -3.40 eV

for n = 3

$E_{n} = 13.6 * 1^{2} / 3^{2}$ eV

= -1.51 eV

for n = 4

$E_{n} = 13.6 * 1^{2} / 4^{2}$ eV

= -0.85 eV

for n = 5

$E_{n} = 13.6 * 1^{2} / 5^{2}$ eV

= -0.544 eV

n = 2 to n = 4 (absorption)

ΔE = E4 - E2 = -0.85 - (-3.40) = 2.55 eV

n = 2 to n = 1 (emission)

ΔE = E1 - E2 = -13.6 - (-3.40) = -10.2eV

The negative sign indicates that emission will take place.

n = 2 to n = 5 (absorption)

ΔE = E5 - E2 = -0.544 - (-3.40) = 2.856 eV

n = 4 to n = 3 (emission)

ΔE = E3 - E4 = -1.51 - (-0.85) = -0.66 eV

We know that

E = h * υ

Therefore, Energy is proportional to frequency.

So increasing the order of energy is

E4 < E1 < E3 < E2

order of frequency is

d < a < c < b

For more information click on the link below:

brainly.com/question/17058029

# SPJ4

8 0

1 year ago

A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.100 moles

alexgriva [62]

Answer : The pH of the solution is, 3.41

Explanation :

First we have to calculate the moles of $HF$ .

$\text{Moles of HF}=\text{Concentration of HF}\times \text{Volume of solution}$

$\text{Moles of HF}=0.250M\times 1.50L=0.375mol$

Now we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (6.8\times 10^{-4})$

$pK_a=4-\log (6.8)$

$pK_a=3.17$

The reaction will be:

$HF+OH^-\rightleftharpoons F^-+H_2O$

Initial moles 0.375 0.100 0.375

At eqm. (0.375-0.100) 0 (0.375+0.100)

= 0.275 = 0.475

Now we have to calculate the pH of solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[F^-]}{[HF]}$

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

$pH=3.17+\log [\frac{(\frac{0.475}{1.50})}{(\frac{0.275}{1.50})}]$

$pH=3.41$

Thus, the pH of the solution is, 3.41

8 0

3 years ago