All categories

3 years ago

Place the elements below in order of decreasing ionization energy. Aluminum(Al) Chlorine(Cl) Magnesium (Mg) Sulfur(S)

Chemistry

1 answer:

arlik [135]3 years ago

7 0

Chlorine
Magnesium
Aluminum
Sulfur

You might be interested in

Radiation that changes direction as it passes through material is called what

RSB [31]

Light? Look up the Third law.
Heat?
Gosh this has been awhile. Please recheck this.

7 0

3 years ago

What is PO2 in prefix form?

Dmitriy789 [7]

Answer:

i am not sure sorry

Explanation:

6 0

3 years ago

How many liters of nitrogen gas is produced if 50.0L of water is produced at STP?

Alona [7]

Answer:

25.0 liter of nitrogen gas

Explanation:

1. Chemical equation

Ammonium nitrite is a solid compound that decomposes into nitrogen gas and water vapor as per this chemical equation:

$NH_4NO_2(s)\rightarrow N_2(g)+2H_2O(g)$

2. Mole ratio

$1molN_2(g)/2molH_2O(g)$

3. Volume ratio

Since, both species are gases and are at same temperature and pressure, the volume ratio is equal to the mol ratio.

Thus, the volume ratio is:

$1literN_2(g)/2literH_2O(g)$

4. Use the volume ratio with the known amount of water produced

$50.0literH_2O(g)\times 1literN_2(g)/2literH_2O(g)=25.0literN_2(g)$

8 0

3 years ago

This is an impossible formula. Pretend it is real.

guapka [62]

There are eight moles of oxygen atoms in 1 mole of $Mn_{9}(ClO_{4}) _{2}$ .

<h3>What is the number of moles of oxygen atoms?</h3>

We know that a compound is composed of atoms. The atoms that make up the molecule are chemically combined. It is usual that the number of atoms in the compound would correspond with the chemical formula.

Now we have the compound $Mn_{9}(ClO_{4}) _{2}$ . In one mole of the compound we have;

9 Moles of manganese atom
2 moles of chlorine atom
8 moles of oxygen atom

Thus, there are eight moles of oxygen atoms in 1 mole of $Mn_{9}(ClO_{4}) _{2}$ .

Learn more about atoms;brainly.com/question/1566330

#SPJ1

7 0

2 years ago

Calculate ΔH o rxn for the following: CH4(g) + Cl2(g) → CCl4(l) + HCl(g)[unbalanced] ΔH o f [CH4(g)] = −74.87 kJ/mol ΔH o f [CCl

anzhelika [568]

Answer: ΔH for the reaction is -277.4 kJ

Explanation:

The balanced chemical reaction is,

$CH_4(g)+Cl_2(g)\rightarrow CCl_4(l)+HCl(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]$

$\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]$

where,

n = number of moles

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

$\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]$

$\Delta H=-277.4kJ$

Therefore, the enthalpy change for this reaction is, -277.4 kJ

4 0

3 years ago