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3 years ago

A neutralization reaction is taking place between solutions of phosphoric acid and calcium

Chemistry

1 answer:

Harrizon [31]3 years ago

4 0

Answer:

Explanation:

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The equilibrium constant, Kc, for the following reaction is 3.61×10-4 at 426 K. PCl5(g) PCl3(g) + Cl2(g) When a sufficiently lar

iogann1982 [59]

Answer:

the concentration of PCl5 in the equilibrium mixture = 296.20M

Explanation:

The concept of equilibrium constant was applied where the equilibrium constant is the ration of the concentration of the product over the concentration of the reactants raised to the power of their coefficients. it can be in terms of concentration in M or in terms of Pressure in atm.

The detaied steps is as shown in the attached file.

8 0

2 years ago

Fumaric acid, which occurs in many plants, contains, by mass, 41.4% carbon, 3.47% hydrogen, and 55.1% oxygen. The molecular mass

lukranit [14]

Answer:

Explanation:

C = 41.4/12 = 3.43

H = 3.47/1 = 3.47

O = 55.1/16 =3.44

CHO is the skeletal formula (divide each by the lowest number above). The results are close enough to 1 to be 1.

(CHO)_x = 116

C + H + O = 29

(29) _ x = 116

x = 116/29

x = 4

So there area 4 carbons 4 hydrogens and 4 oxygens.

The correct formula is C4H4O4

3 0

2 years ago

How many hydrogen atoms are in 709 grams of water? Answer in units of atoms.

zimovet [89]

Answer:

26 Hydrogen atoms

Explanation:

H2O

Each hydrogen atom: 2+16 = 18g

Hence,

1 atom -> 18g

x atoms -> 709g

709/18 = 39 atoms

Therefore, 39 atoms give 709g

Hence, 26 Hydrogen atoms are used

<em>Feel free to mark it as brainliest :D</em>

4 0

2 years ago

Calcium hydroxide, Ca(OH)2, is an ionic compound with a solubility product constant, Ksp, of 6.5×10–6. Calculate the solubility

kari74 [83]

Answer: The solubility of this compound in pure water is 0.012 M

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$

The equation for the ionization of the is given as:

$Ca(OH)_2\rightarrow Ca^{2+}+2OH^-$

By stoichiometry of the reaction:

1 mole of $Ca(OH)_2$ gives 1 mole of $Ca^{2+}$ and 2 mole of $OH^-$

When the solubility of $Ca(OH)_2$ is S moles/liter, then the solubility of $Ca^{2+}$ will be S moles\liter and solubility of $OH^-$ will be 2S moles/liter.

$K_{sp}=[Ca^{2+}][OH^{-}]^2$