All categories

3 years ago

Predict the effect of an eightfold pressure increase on the equilibriunm composition of the reaction 3 Nalg) + H21g) =2 NaHlg)

2 answers:

6 0

According to Le cha telier's principle when we increase the pressure of a equilibrium reaction the reaction shift to the side where few moles of gas present.

Your reaction (Notice - I guess the reaction you written is wrong, but still I'm solving with your given prediction)

3NaI (g) + H2 (g) = 2NaHI (g)

Where access of sodium iodide is reacting with Hydrogen gas to form NaHI molecule.

Number of moles of gas on reaction side - 3+2 = 5 moles

Number of moles of gas on product side - 2 moles

Conclusion- the reaction will shift to the right of the reaction

Assoli18 [71]3 years ago

4 0

According to Le cha telier's principle when we increase the pressure of a equilibrium reaction the reaction shift to the side where few moles of gas present.

Your reaction (Notice - I guess the reaction you written is wrong, but still I'm solving with your given prediction)

3NaI (g) + H2 (g) = 2NaHI (g)

Where access of sodium iodide is reacting with Hydrogen gas to form NaHI molecule.

Number of moles of gas on reaction side - 3+2 = 5 moles

Number of moles of gas on product side - 2 moles

Conclusion- the reaction will shift to the right of the reaction

You might be interested in

Determine the molarity of a 6.0 mole% sulfuric acid solution with SG-a 1.07 Note: Atomic Weight: S (32), O 16); H (O)

marin [14]

Answer:

The molarity of a 6.0 mole% sulfuric acid solution is 2.8157 Molar.

Explanation:

Suppose there are 100 moles in solution:

Moles of sulfuric acid = 6% of 100 moles = 6 moles

Mass of 6 moles of sulfuric acid = 6 mol × 98 g/mol=588 g

Moles of water = 100%- 6% = 94%= 94 moles

Mass of water = 94 mol × 18 g/mol = 1692 g

Specific gravity of the solution ,S.G= 1.07

Density of solution = D

$S.G=\frac{D}{d_w}$

$d_w$ = density of water = 1 g/mL

$D=S.G\times d_w=1.07\times 1 g/mL=1.07 g/mL$

Mass of the solution = 588 g + 1692 g = 2280 g

Volume of the solution = V

Volume = $\frac{Mass}{Density}$

$=\frac{2280 g}{1.07 g/mL}=2130.84 mL=2.13084 L$

1 mL = 0.001 L

$Molarity = \frac{n}{V(L)}$

n = number of moles of compound

V = volume of the solution in L

here we have ,n = 6 moles of sulfuric acid

V = 2.13084 L

So, the molarity of the solution is :

$Molarity=\frac{6 mol}{2.13084 L}=2.8157 mol/L$

5 0

2 years ago

Write the ground-state electron configuration for the chloride ion.You may write either the full or condensed electron configura

dalvyx [7]

Answer:

$1s^22s^22p^63s^23p^6$

Explanation:

Chlorine is the element of group 17 and third period. The atomic number of chlorine is 17 and the symbol of the element is Cl.

The electronic configuration of the element chlorine is:-

$1s^22s^22p^63s^23p^5$

Chloride ion is formed when chlorine atom gain one more electron. So, the ground-state electron configuration for the chloride ion is:-

$1s^22s^22p^63s^23p^6$

8 0

3 years ago

gulaghasi [49]

163 lb * 1 kg / 2.205 lb * 15.0 mg/kg = 1108.8 mg or about 1.11 g

6 0

2 years ago

This graph shows two curves pertaining to a hydrogen s orbital.

fgiga [73]

Answer 1) : According to the complete question attached in the answer,

The radial wave function which is denoted by $R_{nl}(r)$ shown with orange color crosses through zero point. Also, At the the radial nodes, which are spherical shells to some radial distance away from the nucleus there no electron are found.

Also, the radial probability distribution curve denoted as $R^{2}_{nl}(r)$ shown in blue color is observed to touch zero, and shows the place of radial node.

Therefore, the total number of nodes will include both the kinds which has radial and angular nodes which will be represented by 'n'.

It is observed that for any atomic orbital, the total number of nodes will be n-1 .

Considering the s orbital of the hydrogen, which has zero angular momentum (l); (l=0), as it has zero angular nodes.

Hence, there will be only radial nodes, which is

(n−1 = total number of radial nodes in s orbitals)

According to the image, there are 4 radial nodes shown, so n = 5 (as n-1 = 4; therefore, n = 5)

This represents the 5s orbital.

Answer 2) The radial nodes are observed in I'm seeing radial nodes at

$1.9a_{0}$ , $6.4a_{0}$ , $13.9a_{0}$ and $27.0a_{0}$ .

where $a_{0}$ represents the hydorgen bohr atomic radius = 0.0529177 nm

Explanation : It is quite easy to observe the given graph and find out the approximate values of the radial nodes, it does not requires any equation to be solved. Equation can be used to find the radial nodes if it was supplied along with the question. Although by mere speculation one can find out the answer.

3 0

2 years ago

Read 2 more answers

Determina el grado de pureza de un marmol (CaCO3), si al descomponerse 125 g del mismo se desprenden 20 litros de dióxido de car

jarptica [38.1K]

Answer:

67.8%

Explanation:

La reacción de descomposición del CaCO₃ es:

CaCO₃ → CO₂ + CaO

Donde 1 mol de CaCO₃ al descomponerse produce 1 mol de CO₂ y 1 mol de CaO.

Usando la ley general de los gases, las moles de dioxido de carbono son:

PV = nRT.

Donde P es presión (1atm), V es volumen (20L), n son moles de gas, R es la constante de los gases (0.082atmL/molK) y T es temperatura absoluta (15 + 273.15 = 288.15K). Reemplazando los valores en la ecuación:

PV / RT = n

1atmₓ20L / 0.082atmL/molKₓ288.15K = 0.846 moles

Como 1 mol de CO₂ es producido desde 1 mol de CaCO₃, las moles iniciales de CaCO₃ son 0.846moles.

La masa molar de CaCO₃ es 100.087g/mol. Así, la masa de 0.846moles de CaCO₃ es:

0.846moles ₓ (100.087g / mol) = 84.7g de CaCO₃

Así, la pureza del marmol es:

(84.7g de CaCO₃ / 125g) ₓ 100 =

7 0

2 years ago

Predict the effect of an eightfold pressure increase on the equilibriunm composition of the reaction 3 Nalg) + H21g) =2 NaHlg)​

Predict the effect of an eightfold pressure increase on the equilibriunm composition of the reaction 3 Nalg) + H21g) =2 NaHlg)