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Complete and balance the precipitation reactions. Include physical states. Refer to the solubility rules as necessary. precipita

Tcecarenko [31]

The balanced chemical reactions are:

$AgNO{_3}(aq) \ + \ NaCl(aq) \rightarrow \ AgCl(s) \ + NaNO_{3}(aq)\\2K_{3}PO_{4}(aq) \ + 3MgCl_{2}(aq) \rightarrow \ 6KCl(aq) \ + Mg_{3}(PO_{4})_{2}(s)$

Further Explanation:

The following reactions will undergo double displacement where the metal cations in each compound are exchanged and form new products.

For reaction 1, the compounds involved are nitrates and chlorides. To determine the states of the products, the solubility rules for nitrates and chlorides must be followed:

All nitrates are generally soluble.
Chlorides are generally soluble except AgCl, PbCl2, and Hg2Cl2.

Therefore, the products will have the following characteristics:

silver chloride (AgCl) is insoluble
sodium nitrate (NaNO3) is soluble

For reaction 2, the compounds involved are phosphates and chlorides. The solubility rules for phosphates and chlorides are as follows:

Phosphates are generally insoluble except for Group 1 phosphates.
Chlorides are generally soluble except for AgCl, PbCl2, and Hg2Cl2.

Hence, the products of the second reaction will have the following characteristics:

potassium chloride (KCl) is soluble
magnesium phosphate is insoluble

Insoluble substances are denoted by the symbols (s) in a chemical equation. The soluble substances are denoted as (aq).

Learn More

Learn more about solubility brainly.com/question/1163248
Learn more about solubility rules brainly.com/question/1447161
Learn more about net ionic equations brainly.com/question/12980075

Keywords: solubility rules, precipitation reaction

3 0

3 years ago

Help please its science

Nookie1986 [14]

Answer: any answer choices

Explanation:

6 0

3 years ago

The ionic radius for Na+ is 0.097 ηm and for Cl- is 0.181 ηm, the absolute value of the charge for each ion is 1.6x10-19 C, ε_o=

Vanyuwa [196]

Answer:

B = (2.953 × 10⁻⁹⁵) N.m⁹

Explanation:

At equilibrium, where the distance between the two ions (ro) is the sum of their ionic radii, the force between the two ions is zero.

That is,

Fa + Fr = 0

Fa = - Fr

Fa = (|q₁q₂|)/(4πε₀r²)

Fr = -B/(r^n) but n = 9

Fr = -B/r⁹

(|q₁q₂|)/(4πε₀r²) = (B/r⁹)

|q₁| = |q₂| = (1.6 × 10⁻¹⁹) C

(1/4πε₀) = k = (8.99 × 10⁹) Nm²/C²

r = 0.097 + 0.181 = 0.278 nm = (2.78 × 10⁻¹⁰) m

(k|q₁q₂|)/(r²) = (B/r⁹)

(k × |q₁q₂|) = (B/r⁷)

B = (k × |q₁q₂| × r⁷)

B = [8.99 × 10⁹ × 1.6 × 10⁻¹⁹ × 1.6 × 10⁻¹⁹ × (2.78 × 10⁻¹⁰)⁷]

B = (2.953 × 10⁻⁹⁵) N.m⁹

6 0

3 years ago

Bilangan oksidasi vanadium paling tinggi terdapat dalam senyawa..

emmasim [6.3K]

The answer is (b). As, vanadium is attached to five fluoride atoms, each flouride containing -1 oxidation state, hence five fluoride contains -5, to neutralize, vanadium should have +5 oxidation state.

8 0

3 years ago

What volume of benzene (C6H6, d= 0.88 g/mL, molar mass = 78.11 g/mol) is required to produce 1.5 x 103 kJ of heat according to t

Anton [14]

Answer:

We need 42.4 mL of benzene to produce 1.5 *10³ kJ of heat

Explanation:

Step 1: Data given

Density of benzene = 0.88 g/mL

Molar mass of benzene = 78.11 g/mol

Heat produced = 1.5 * 10³ kJ

Step 2: The balanced equation

2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔH°rxn = -6278 kJ

Step 3: Calculate moles of benzen

1.5 * 10³ kJ * (2 mol C6H6 / 6278 kJ) = 0.478 mol C6H6

Step 4: Calculate mass of benzene

Mass benzene : moles benzene * molar mass benzene

Mass benzene= 0.478 * 78.11 g

Mass of benzene = 37.34 grams

Step 5: Calculate volume of benzene

Volume benzene = mass / density

Volume benzene = 37.34 grams / 0.88g/mL

Volume benzene = 42.4 mL

We need 42.4 mL of benzene to produce 1.5 *10³ kJ of heat

4 0

3 years ago