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Kamila [148]
2 years ago
8

The amount of gas that occupies 36.52 L at 68.0°C and 672 mm Hg is __________ mol.

Chemistry
1 answer:
ASHA 777 [7]2 years ago
7 0
We assume that this gas is an ideal gas. We use the ideal gas equation to calculate the amount of the gas in moles. It is expressed as:

PV = nRT
(672) (1/760) (36.52) = n (0.08206) ( 68 +273.15)
n = 1.15 mol of gas

Hope this answers the question. Have a nice day.
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Aluminum oxide reacts with hydrochloric acid to produce salt. What is the name of the salt produced?
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IT IS CHLORIDE!

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B. It shifts the equilibrium toward the right, favoring product.
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11) the difference in heat energies between products and reactants

12) enthalpy change

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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
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the concentration of PCl5 in the equilibrium mixture = 296.20M

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The detaied steps is as shown in the attached file.

8 0
2 years ago
Hydrogen iodide is not produced by the same method is for hydrogen chloride why with reaction​
Masja [62]

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H2SO4 + NaCl ==> HCl + NaHSO4

This method(Using H2So4) will work for all hydrogen hydrogen halide except Hydrogen Iodide and Hydrogen Bromide.

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5 0
2 years ago
When aqueous solutions of manganese(II) iodide and sodium phosphate are combined, solid manganese(II) phosphate and a solution o
Feliz [49]

Answer:

The net ionic equation for the given reaction :

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Explanation:

3MnI_2(aq)+2Na_3PO_4_2(aq)\rightarrow Mn_3(PO_4)_2(s)+6NaI(aq)...[1]

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Na_3PO_4(aq)\rightarrow 3Na^{+}(aq)+PO_4^{3-}(aq)...[3]

NaI(aq)\rightarrow Na^+(aq)+I^-(aq)

Replacing MnI_2(aq) , NaI and Na_3PO_4(aq) in [1] by usig [2] [3] and [4]

3Mn^{2+}(aq)+6I^-(aq)+6Na^{+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)+6Na^+(aq)+6I^-(aq)

Removing the common ions present ion both the sides, we get the net ionic equation for the given reaction [1]:

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

8 0
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