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meriva
2 years ago
6

An element forms an oxide, E₂O₃, and a fluoride, EF₃.(a) Of which two groups might E be a member?

Chemistry
1 answer:
Fittoniya [83]2 years ago
8 0

a) The E might belong to group 13.

As the formula of a chemical compound is derived by the cross multiplication of the valency of the atoms. As formula of the given oxide is  and valency of O atom is -2, therefore valency of element E must be +3 in order to obtain E2O3.

Also, in EF3, the valency of E will be +3 because there are three atoms of fluorine who has an individual valency of -1. Thus, e will have the valency of +3.

The Group 13 is the boron group which has the following elements:

  • Boron
  • Aluminium
  • Gallium
  • Indium
  • Thallium

All these elements have the valency of +3.

To know more about Valency, refer to this link:

brainly.com/question/12717954

#SPJ4

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If 63.5 mol of an ideal gas is at 9.11 atm at 42.80 °C, what is the volume of the gas?
Anastaziya [24]

The volume of the gas is 180.26 L, if there are 63.5 mol of an ideal gas at 9.11 atm at 42.80 °C.

Applying the ideal gas law PV= nRT

After rearranging the aforementioned expression, the volume might then be found as: V= n R T/ P.

Consequently, V= 63.5 mol, 0.0821, 315 K, and 9.11 atm equal 180.26 L.

<h3>How is the ideal gas equation defined?</h3>

The ideal gas equation is PV = nRT. In this equation, P denotes the ideal gas's pressure, V its volume, n its total amount, expressed in moles, and R its resistance for the universal gas constant, and T for temperature.

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3 0
1 year ago
Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid
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pH of resulting solution = 7.98

Explanation:

The balanced equation  

HA + NaOH - Na+ + A- + H2O

Number of moles of A = Number of moles of HA  = Number of moles of NaOH

= 35.8/1000 * 0.020 = 0.000716 mol

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pKb = 14 – pKa = 14 -3.9 = 10.1

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Kb = [HA][OH-]/[A-]

Kb = a^2/(0.01178 -a) = 7.943 * 10^-11

a^2 + 7.943 * 10^-11 a – 9.357 * 10^-13 = 0

a = 9.673 * 10^-7

OH- = a = 9.673 * 10^-7 M

pOH = -log [OH-] = -log (9.673 * 10^-7) = 6.02

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