Mrs. Rushing fills a balloon with hydrogen gas to demonstrate its ability to burn. Which combination could she

7.47 Two atoms have the electron configurations 1s22s22p6 and 1s22s22p63s1. The first ionization energy of one is 2080 kJ/mol an

Agata [3.3K]

Answer:

2080 kJ/mol is the first ionization of 1st atom and 496 kJ/mol is the first ionization of 2nd atom

Explanation:

Given electronic configurations are :

1st: $1s^22s^22p^6$

2nd : $1s^22s^22p^63s^1$

given 1st ionization energy are: 2080 kJ/mol and 496 kJ/mol

generally ionization energy of fulfilled orbital is more than half filled orbital and these two state are more stable.

therefore ionization energy of fulfilled is more than half filled orbital

hence

ionization energy of 1st atom will be very high because its orbital is fulfilled and less energy for 2nd atom so 2080 kJ/mol is the first ionization of 1st atom and 496 kJ/mol is the first ionization of 2nd atom.

4 0

2 years ago

Find the time Δt it takes the magnetic field to drop to zero. Express your answer in terms of some or all of the quantities a, B

Mekhanik [1.2K]

Δt= $\frac{NB_0 \pi a^2 }{IR}$

This because ε = dΦ/dt, and emf is a voltage so it can also be written as dΦ/dt = IR

Because ΔΦ= ABcosθ for each of the N loops and cosθ is just 1 here only A needs to be replaced, and it can be replaced with A=πr^2, which makes that part of the equation ΔΦ_m=N*B0*πa^2

7 0

2 years ago

Suppose that 0.323 g of an unknown sulfate salt is dissolved in 50 mL of water. The solution is acidified with 6M HCl, heated, a

geniusboy [140]

Answer:

1) 41.16 % = 0.182 grams

2) The alkali cation is K+ , to form the salt K2SO4

Explanation:

Step 1: Data given

Mass of unknown sulfate salt = 0.323 grams

Volume of water = 50 mL

Molarity of HCl = 6M

Step 2: The balanced equation

SO4^2- + BaCl2 → BaSO4 + 2Cl-

Step 3: Calculate amount of SO4^2- in BaSO4

The precipitate will be BaSO4

The amount of SO4^2- in BaSO4 = (Molar mass of SO4^2-/Molar mass BaSO4)*100 %

The amount of SO4^2- in BaSO4 = (96.06 /233.38) * 100

= 41.16%

So in 0.443g of BaSO4 there will be 0.443 * 41.16 % = <u>0.182 grams</u>

<u />

2. If it is assumed that the salt is an alkali sulfate determine the identity of the alkali cation.

The unknown sulphate salt has 0.182g of sulphate. This means the alkali cation has a weight of 0.323-0.182 = 0.141g grams

An alkali cation has a chargoe of +1; sulphate has a charge of -2

The formula will be X2SO4 (with X = the unknown alkali metal).

Calculate moles of sulphate

Moles sulphate = 0.182 grams (32.1 + 4*16)

Moles sulphate = 0.00189 moles

The moles of sulphate = 0.182/(32.1+16*4)

The moles of sulphate = 0.00189 moles

X2SO4 → 2X+ + SO4^2-

For 2 moles cation we have 1 mol anion

For 0.00189 moles anion, we have 2*0.00189 = 0.00378 moles cation

Calculate molar mass

Molar mass = mass / moles

Molar mass = 0.141 grams / 0.00378 grams

Molar mass = 37.3 g/mol

The closest alkali metal is potassium. (K2SO4 )

3 0

2 years ago

2. Suggest four ways in which the concentration of PH3 could be increased in an equilibrium described by the following equation:

Nesterboy [21]

Answer

increase in temperature

decrease in pressure

continuous removal of PH3
adding more of P into the system

Explanation:

In the reaction P4(g)+6H2(g) ⇌ 4PH3(g);

The effect of temperature on equilibrium has to do with the heat of reaction. Recall that for an endothermic reaction, heat is absorbed in the reaction, and the value of ΔH is positive. Thus, for an endothermic reaction, we can picture heat as being a reactant:

heat+A⇌BΔH=+

Since the reaction is endothermic reaction, heat is a absorbed. Decreasing the temperature will shift the equilibrium to the left, while increasing the temperature will shift the equilibrium to the right forming more of PH3.

According to Le Chatelier’s principle, adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products. In the same Way, reducing the concentration of the product will also shift equilibrium to the right continually forming PH3 as it is removed.

4 0

2 years ago

Now consider the reaction when 45.0 g NaOH have been added. What amount of NaOH is this, and what amount of FeCl3 can be consume

gayaneshka [121]

The correct answer is that 1.125 mol of NaOH is available, and 60.75 g of FeCl₃ can be consumed.

The mass of NaOH is 45 g

The molar mass of NaOH = 40 g/mol

The moles of NaOH = mass / molar mass

= 45 / 40

= 1.125

Thus, 1.125 mol NaOH is available

3 NaOH + FeCl₃ ⇒ Fe (OH)₃ + 3NaCl

3 mol of NaOH react with 1 mol of FeCl₃

1.125 moles of NaOH will react with x moles of FeCl₃

x = 1.125 / 3

x = 0.375 mol

0.375 mol FeCl₃ can take part in reaction

The molar mass of FeCl₃ is 162 g/mol

The mass of FeCl₃ = moles × mass

= 0.375 × 162

= 60.75 g

Thus, the amount of FeCl₃, which can be consumed is 60.75 g

6 0

2 years ago