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Mkey [24]
3 years ago
10

Help!!!

Chemistry
2 answers:
Maru [420]3 years ago
5 0
45.) 3 Ca(OH)2(aq) + 2 H3Po4(aq) = 6 H2O(l) + Ca3(Po4)2(s)

I'll try to find the others.
Maurinko [17]3 years ago
3 0

45) Balanced chemical reaction:

3Ca(OH)₂(aq) + 2H₃PO₄(aq) → Ca₃(PO₄)₂(aq) + 6H₂O(l).

Balanced ionic equation:

3Ca²⁺(aq) + 6OH⁻(aq) + 6H⁺(aq) + 2PO₄³⁻(aq) → 3Ca²⁺(aq) + 2PO₄³⁻(aq)+ 6H₂O(l).

Balanced net ionic equation: 6OH⁻(aq) + 6H⁺(aq) → 6H₂O(l).

This reaction is example of neutralization reaction.

46) Answer is: 67.2 liters of oxygen are needed to react completely.

Balanced chemic equation: 2H₂S(g) + 3O₂(g) → 2H₂O(l) + 2SO₂(g).

V(H₂S) = 45.0 L; volume of hydrogen sulfide.

Vm = 22.4 L/mol; molar volume at STP.

n(H₂S) = 45 L ÷ 22.4 L/mol.

n(H₂S) = 2.01 mol; amount of hydrogen sulfide.

From balanced chemical reaction: n(H₂S) : n(O₂) = 2 : 3.

n(O₂) = 3 mol; amount of oxygen.

V(O₂) = 3 mol · 22.4 L/mol.

V(O₂) = 67.2 L; volume of oxygen.

47) The azimuthal quantum number (l) determines its orbital angular momentum and describes the shape of the orbital.  

s-orbitals (for example 1s, 2s) are spherically symmetric around the nucleus of the atom.

p-orbitals are dumb-bell shaped. l = 0,1...n-1, when l = 1, that is p subshell.

d-orbitals are butterfly shaped.

f-orbitals are similar to the d orbital, but cut in half (eight lobes instead of four).

48) Answer is: Ionic bond is the electrostatic attraction between oppositely charged ions (cations and anions).

For example, compound magnesium chloride (MgCl₂) has ionic bond.

Magnesium (Mg) transfers two electrons, and form positive cation Mg²⁺, to two chlorine atoms (Cl), which form negative anions Cl⁻.

Magnesium is metal from group 2 and chlorine is nonmetal from group 17 of the Periodic table.

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Ten kilograms of R-134a fill a 1.595-m3 weighted piston-cylinder device at a temperature of −26.4°C. The container is now heated
Tju [1.3M]

Explanation:

The given data is as follows.

    Mass of refrigerant, m = 10 kg

  Volume of the refrigerant, V = 1.595 m^{3}

Formula for specific volume of the refrigerant is as follows.

        v = \frac{V}{m}

           = 0.1595 m^{3}/kg

So, at -26.4^{o}C specific volume will be within v_{f} and v_{g} and pressure is constant.

The fluid will be in super-heated state at temperature 100^{o}C and at T = -26.4^{o}C pressure 1 bar = 0.1 MPa.

According to super-heated tables, the specific volume is v = 0.30138 m^{3}/kg.

Hence, the final volume will be calculated as follows.

               V_{f} = v \times m

                         = 0.30138 m^{3}/kg \times 10 kg

                         = 3.0138 m^{3}

Thus, we can conclude that final volume of the R-134a is 3.0138 m^{3}.

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