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____
1 month ago
14

Notice that "SO4" appears in two different places in this chemical equation. SO42− is a polyatomic ion called "sulfate." What nu

mber should be placed in front of CaSO4 to give the same total number of sulfate ions on each side of the equation? ?CaSO4+AlCl3→CaCl2+Al2(SO4)3
Chemistry
1 answer:
VMariaS [2.9K]1 month ago
5 0
The coefficient that should precede CuSO_4 is three (3). Explanation: A balanced chemical equation requires that the number of atoms for each element in the reactants must equate to those in the products. The provided unbalanced chemical equation is noted, revealing an imbalance between the number of chloride and sulfate ions. To rectify this, coefficients of 3, 2, and 3 are used in the respective places to achieve balance, resulting in the appropriately adjusted equation.
You might be interested in
Which of these are physical changes in matter? Check all that apply boiling water a pencil being sharpened exploding dynamite fr
Anarel [2989]

Explanation:

A physical change refers to a modification that does not alter the chemical makeup of a substance.

For instance, characteristics such as shape, size, mass, volume, density, and boiling point are considered physical properties.

Conversely, alterations that result in a change in the chemical structure of a substance are classified as chemical changes.

Examples include combustible substances like dynamite or decomposing cheese, which represent chemical transformations.

Thus, we can conclude that the following are all physical changes in matter:

  • boiling water.
  • a pencil being sharpened.
  • freezing water.
3 0
1 month ago
Read 2 more answers
A sample of solid naphthalene is introduced into an evacuated flask. Use the data below to calculate the equilibrium vapor press
Tems11 [2777]

Answer: The vapor pressure of naphthalene within the flask remains at 2.906\times 10^{-4} atm.

Explanation:

The transformation from solid naphthalene to its gaseous form follows the equilibrium reaction:

C_{10}H_8(s)\rightleftharpoons C_{10}H_8(g)

  • The formula employed to determine the enthalpy change for the reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]

The formula for calculating the enthalpy change regarding the aforementioned reaction is:

\Delta H^o_{rxn}=(1\times \Delta H^o_f_{(C_{10}H_8(g))})-(1\times \Delta H^o_f_{(C_{10}H_8(s))})

The provided information includes:

\Delta H^o_f_{(C_{10}H_8(s))}=78.5kJ/mol\\\Delta H^o_f_{(C_{10}H_8(g))}=150.6kJ/mol

Substituting the values into the previous equation produces:

\Delta H^o_{rxn}=(1\times 150.6)-(1\times 78.5)=72.1kJ/mol

  • The formula utilized to compute Gibbs free energy change is of a reaction:

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f(product)]-\sum [n\times \Delta G^o_f(reactant)]

The equation for the enthalpy change for the reaction is:

\Delta G^o_{rxn}=(1\times \Delta G^o_f_{(C_{10}H_8(g))})-(1\times \Delta G^o_f_{(C_{10}H_8(s))})

The given factors include:

\Delta G^o_f_{(C_{10}H_8(s))}=201.6kJ/mol\\\Delta G^o_f_{(C_{10}H_8(g))}=224.1kJ/mol

By inserting values from the above equation, we arrive at:

\Delta G^o_{rxn}=(1\times 224.1)-(1\times 201.6)=22.5kJ/mol

  • For the calculation of K_1 (at 25°C) regarding the provided value of Gibbs free energy, the following relationship is applied:

\Delta G^o=-RT\ln K_1

where,

\Delta G^o = Gibbs free energy = 22.5 kJ/mol = 22500 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = 8.314J/K mol

T = temperature = 25^oC=[273+25]K=298K

K_1 = equilibrium constant at 25°C =?

Inserting values into the above equation yields:

22500J/mol=-(8.314J/Kmol)\times 298K\times \ln K_1\\\\K_1=1.14\times 10^{-4}

  • To determine the equilibrium constant at 35°C, we refer to the equation proposed by Arrhenius, which states:

\ln(\frac{K_2}{K_1})=\frac{\Delta H}{T}(\frac{1}{T_1}-\frac{1}{T_2})

where,

K_2 = Equilibrium constant at 35°C =?

K_1 = Equilibrium constant at 25°C = 1.14\times 10^{-4}

\Delta H = Enthalpy change of the reaction = 72.1 kJ/mol = 72100 J

R = Gas constant = 8.314J/K mol

T_1 = Initial temperature = 25^oC=[273+25]K=298K

T_2 = Final temperature = 35^oC=[273+35]K=308K

By plugging values into the equation above, we obtain:

\ln(\frac{K_2}{1.14\times 10^{-4}})=\frac{72100J/mol}{8.314J/K.mol}(\frac{1}{298}-\frac{1}{308})\\\\K_2=2.906\times 10^{-4}

  • In order to calculate the partial pressure of naphthalene at 35°C, we utilize the equation for K_p, which is:

K_p=\frac{p_{C_{10}H_8(g)}}{p_{C_{10}H_8(g)}}=p_{C_{10}H_8(g)

The partial pressure of the solid phase is considered to be 1 at equilibrium.

Therefore, the value for K_2 will equal K_p

p_{C_{10}H_8}=2.906\times 10^{-4}

Consequently, the partial pressure of naphthalene at 35°C is 2.906\times 10^{-4} atm.

3 0
1 month ago
The atomic radius of magnesium is 150 pm. The atomic radius of strontium is 200 pm. What is the atomic radius of calcium
castortr0y [3046]

Answer:

Calcium's atomic radius is roughly 175 pm.

Explanation:

We know that magnesium has an atomic radius of 150 pm.

The atomic radius of strontium measures 200 pm.

Since calcium's position is between magnesium and strontium in group 2 of the periodic table, its atomic radius should be roughly averaged between magnesium's and strontium's atomic radii because atomic radius is not constant.

Thus;

Calcium's atomic radius is approximately calculated as follows;

The average atomic radius is (200 + 150)/2 = 175 pm.

7 0
3 months ago
Calculate the number of C atoms in 9.837 x 1024 molecules of CO2. <br><br>Please help
alisha [2963]

Response:

Each molecule consists of one carbon atom.

9.837 * 10^24 molecules correspond to 9.837 * 10^24 carbon atoms.

It's a 1:1 relationship.

6 0
1 month ago
Why are salt and sugar both able to dissolve in water, even though the solutes have different types of chemical bonding?
castortr0y [3046]

Response: Water is a polar substance, facilitating the dissolution of ionic compounds due to the principle that similar types mix.

Ionic interactions occur between salt and water

Sugar contains hydroxyl groups that can form hydrogen bonds with water molecules.

[Hydrogen bond: this refers to the attraction between a hydrogen atom bonded to a highly electronegative atom (like F, O, or N) and another highly electronegative atom (F, O, or N)]

Thus, due to the presence of hydrogen bonds, sugar dissolves in water.

Clarification: Water molecules are polar, exhibiting partial positive charges on the hydrogen atoms and a partial negative charge on the oxygen atom. This allows them to interact with ionic compounds such as salt (NaCl). These interactions occur through the partial charges on water, which attract opposite charges. When dissolved in water, NaCl dissociates into sodium and chloride ions; sodium ions are surrounded by negatively charged oxygen from water, while chloride ions are surrounded by positively charged hydrogens from water. As a result, salt dissolves in water.

Sugar, being a covalent compound, has bonds where electrons are shared unevenly, creating slight positive and negative charges. This characteristic allows sugar to interact with the polar ends of water, facilitating its dissolution. Therefore, it can be stated that sugar dissolves in water due to both substances being polar.

In summary, water is capable of dissolving most polar or ionic substances, as seen with sugar and salt.

6 0
2 months ago
Read 2 more answers
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