Consider the volume of gas collected in the trial in Experiment 2 for Ni at

Element X has the following valence electron configuration: [core]ns2np5. Element M has the following valence electron configura

alex41 [277]

Answer:

The kind of ionic compound formed is MX2.

Explanation:

Element X electron configuration is represented as [core] ns2np5. The group in the periodic table this element belong to is group 7A. The element group is called the halogen family. Element X cannot be stated specifically, because the number is represented with n. Element X will behave as an anions when it react with a metal(cations). Element X has a charge of -1. The element X will gain electron when it bond with a metal. Element X is a non metal . Elements in this group are fluorine, chlorine, bromine, iodine , astatine, and tennessine . The element X have 7 valency electrons.

Element M electronic configuration is represented as [core]ns2. The group in the periodic table this element belong to is group 2A . The element group is called the alkaline earth metals family . Element M will behave as a cation when it bond with a non metal. Element M is a metal , therefore it will likely lose electron to form cations during bonding . The charge of element M is 2+. Element M is positively charged. Elements that belong to this group includes beryllium, magnesium, calcium, strontium, barium and radium. Element M has 2 valency electrons.

The reaction between this 2 ions will likely form an ionic compound . The element M is the cations while the element X is the anions. The element M will lose 2 electron while 2 atoms of element X will gain 2 electrons.Element M will lose 2 electron to attain a stable configuration while 2 atoms of element X will gain a single electron each to attain a stable configuration.

M²+ and F- . This will form MX2 when you cross multiply the charge. The kind of ionic compound formed is MX2.

6 0

3 years ago

At what temp will a gas be at if you allow it to expand from an original 456 mL to 65°C to 3.4 L

alexira [117]

We use the gas law named Charle's law for the calculation of the second temperature. The law states that,
V₁T₂ = V₂T₁
Substituting the known values,
(0.456 L)(65 + 273.15) = (3.4 L)(T₁)
T₁ = 45.33 K

5 0

3 years ago

Read 2 more answers

Mg(OH)2 + HCl --> MgCl2 + H2O

Rashid [163]

I am guessing you want us to balance this equation so.
To balance, we add another molecule of HCl to the left side of the equation and another molecule of water (H20) to the right side of the equation to give:

Mg(OH)2 + 2HCl = MgCl2 + 2H20

6 0

3 years ago

What mass of HgO is required to produce 0.692 mol of O2? 2HgO(s) -> 2Hg(l) + O2(g)

Vika [28.1K]

The answer for the following problem is mentioned below.

Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule

Explanation:

Given:

no of moles of the oxygen gas = 0.692

Also given:

2 HgO → 2 Hg + $O_{2}$

where,

HgO represents mercuric oxide

Hg represents mercury

$O_{2}$ represents oxygen

To calculate:

Molar mass of HgO:

Molar mass of HgO = 216 grams

molar mass of mercury (Hg) = 200 grams

molar mass of oxygen (O) =16 grams

HgO = 200 +16 = 216 grams

We know;

2×216 grams of HgO → 1 mole of oxygen molecule

? → 0.692 moles of oxygen molecule

= $\frac{2*216*0.692}{1}$

= 298.944 grams of HgO

Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule

7 0

3 years ago

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You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

Now put all the given values in the above formula, we get:

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

3 years ago