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aliina [53]
2 years ago
14

How many groups are in the modern periodic table?

Chemistry
2 answers:
Jet001 [13]2 years ago
8 0

Answer:

IA....... VIIIA. main groups (A groups)

IB....... VIIIB. sub groups (B groups)

totally they are 18

Ainat [17]2 years ago
5 0

Answer:

there are 18 groups

Explanation:

You might be interested in
Given that cao(s) + h2o(l) → ca(oh)2(s), δh°rxn = –64.8 kj/mol, how many grams of cao must react in order to liberate 525 kj of
USPshnik [31]

Answer:

454.3 g.

Explanation:

  • From the given data:

1.0 mol of CaO liberates → – 64.8 kJ.

??? mol of CaO liberates → - 525  kJ.

∴ The no. of moles needed = (1.0 mol)(- 525 kJ)/(- 64.8 kJ) = 8.1 mol.

<em>∴ The no. of grams of CaO needed = no. of moles x molar mass</em> = (8.1 mol)(56.077 g/mol) = <em>454.3 g.</em>

8 0
3 years ago
Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th
WINSTONCH [101]

<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]

We are given:

\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

6 0
3 years ago
6.5 Moles of Al reacts with 7.2 Moles of H2O What is the limiting reactant and calculate the Theoretical Yield?
SVEN [57.7K]

Answer:

H₂O is the limiting reactant

Theoretical yield of 240 g Al₂O₃ and 14 g H₂

Explanation:

Find how many moles of one reactant is needed to completely react with the other.

6.5 mol Al × (3 mol H₂O / 2 mol Al) = 9.75 mol H₂O

We need 9.75 mol of H₂O to completely react with 6.5 mol of Al.  But we only have 7.2 mol of H₂O.  Therefore, H₂O is the limiting reactant.

Now find the theoretical yield:

7.2 mol H₂O × (1 mol Al₂O₃ / 3 mol H₂O) × (102 g Al₂O₃ / mol Al₂O₃) ≈ 240 g Al₂O₃

7.2 mol H₂O × (3 mol H₂ / 3 mol H₂O) × (2 g H₂ / mol H₂) ≈ 14 g H₂

Since the data was given to two significant figures, we must round our answer to two significant figures as well.

4 0
3 years ago
The combustion of 0.1240 kg of propane in the presence of excess oxygen produces 0.3110 kg of carbon dioxide. What is the limiti
nadezda [96]

Answer:

The limiting reactant is the propane gas, C₃H₈ while the percentage yield is 83.77%

Explanation:

Here we have

Propane gas with molecular formula C₃H₈, molar mass  = 44.1 g/mol combining with O₂ as follows

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Therefore, 1 mole of C₃H₈  combines with 5 moles of O₂ to produce 3 moles CO₂ and 4 moles of H₂O

Mass of propane = 0.1240 kg = 124.0 g

Number of moles of propane = mass of propane/(molar mass of propane)

The number of moles of propane = 124/44.1 = 2.812 moles

The molar mass of CO₂ = 44.01 g/mol

Mass of CO₂ = 0.3110 kg = 311.0 g

Therefore, number of moles of CO₂ = mass of CO₂/(molar mass of CO₂)

The number of moles of CO₂ = 311.0 kg/ 44.01 g/mol = 7.067 moles

Therefore, since 1 mole of propane produces 3 moles of CO₂, 2.812 moles of propane will produce 3 × 2.812 moles or 8.44 moles of CO₂

Therefore;

The limiting reactant is the propane gas, C₃H₈, since the oxygen is in excess

Hence

The \ percentage \ yield = \frac{Actual \, yield}{Theoretical \, yield} \times 100 = \frac{7.067}{8.44} \times 100 = 83.77 \%

The percentage yield = 83.77%.

7 0
3 years ago
Write a balanced chemical equation for the reaction of copper(II) sulfate and concentrated ammonia to produce teramine copper(II
Rudiy27

Answer:

CuSO4(aq) + 4NH3(aq) + nH2O→ [Cu(NH3)4H2On]S04(aq)

Explanation:

When the concentrated ammonia is added into the copper ions solution, a blue solid is formed. This blue solid is gelatinous.it consist of Cu(OH)2 in insoluble form. As more ammonia is added precipitate start to dissolve and [Cu(NH3)4]∧+2 is produce.

Net Ionic equation:

Cu∧+2 +4NH3 →  [Cu(NH3)4]∧+2

S04∧-2 are spectator ions.

Properties and uses of Tetraaminecopper(ll) sulfate:

1. It is solid compound.

2. Its color is dark blue.

3. It has ammonia like odor.

4. It is used to make the copper compounds and also used in the printing of fabric.

5. It is also used as pesticide.

3 0
3 years ago
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