(I will give brainliest to the person who answered my question correctly!)

Ammonium Iodide dissociates reversibly to ammonia and hydrogen iodide:

NeTakaya

Answer:

The partial pressure of ammonia at equilibrium when a sufficient quantity of ammonium iodide is heated to 400°C Is 0.103 atm.

The correct option is A.

Explanation;

NH4I(s) ⇋ NH3(g) + HI(g)Kp = 0.215 at 400°C

NH4I(s)= 0.215

NH3(g)=0.103

HI(g)Kp=0.112

Therefore = 0.103 +0.112= 0.215

Therefore the partial pressure of ammonia at equilibrium is 0.103 atm

7 0

3 years ago

Hi everyone can anyone help with this, the question and diagram is in the pic thx!

seraphim [82]

Answer:

QP

Explanation:

P has 9 electrons.

Electronic Configuration : 2, 7

Valence electrons : 7

P needs 1 electron to get stable electronic configuration.

Q has 3 electrons.

Electronic Configuration : 2, 1

Valence electrons : 1

P needs to loose 1 electron to get stable electronic configuration.

Q donates 1 electron,

Q -----> Q+ + 1 e-

P gains 1 electron,

P + 1 e- -----> P-

Q+ + P- -----> QP

This is an ionic compound.

8 0

3 years ago

1. How many milliliters of 10.0 M HNO 3 are needed to prepare 0.350 L of 0.400 M solution?

tatyana61 [14]

<h3>Answer:</h3>

14 milliliters

<h3>Explanation:</h3>

We are given;

10.0 M HNO₃

Prepared solution;

Volume of solution as 0.350 L
Molarity as 0.40 M

We are required to determine the initial volume of HNO₃

We are going to use the dilution formula;
The dilution formula is;

M₁V₁ = M₂V₂

Rearranging the formula;

V₁ = M₂V₂ ÷ M₁

=(0.40 M × 0.350 L) ÷ 10.0 M

= 0.014 L

But, 1 L = 1000 mL

Therefore,

Volume = 14 mL

Thus, the volume of 10.0 M HNO₃ is 14 mL

5 0

3 years ago

Be sure to answer all parts. Write the equations representing the following processes. In each case, be sure to indicate the phy

barxatty [35]

Answer:

1. $S^{-}(g)+e^{-} \rightarrow S^{2-}(g)$

2. $Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}$

3.The electron affinity of $Mg^{2+}$ is zero.

4. $O^{2-}(g) \rightarrow O^{-}(g)+e^{-}$

Explanation:

1.

<u>Electron affinity:</u>

It is defined as the amount of energy change when an electron is added to atom in the gaseous phase.

The electron affinity of $S^{-}$ is as follows.

$S^{-}(g)+e^{-} \rightarrow S^{2-}(g)$

2.

<u>Ionization energy</u>:

Amount of energy required to removal of an electron from an isolated gaseous atom.

The third ionization energy of Titanium is as follows.

$Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}$

3.

The electronic configuration of Mg: $1s^{2}2s^{2}2p^{6}3s^{2}$

By the removal of two electrons from a magnesium element we get $Mg^{2+}$ ion.

$Mg^{2+}$ has inert gas configuration i.e, $1s^{2}2s^{2}2p^{6}$

Hence, it does not require more electrons to get stability.

Therefore,the electron affinity of $Mg^{2+}$ is zero.

4.

The ionization energy of $O^{2-}$ is follows.

$O^{2-}(g) \rightarrow O^{-}(g)+e^{-}$

3 0

3 years ago

Please help me will mark brainliest

LuckyWell [14K]

Answer:

accretion

Explanation:

the coming together and cohesion of matter under the influence of gravitation to form larger bodies.

8 0

2 years ago