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qaws [65]
3 years ago
9

Why is there a bond between the ions in potassium chloride

Chemistry
1 answer:
yKpoI14uk [10]3 years ago
6 0

Answer:

Explanation:

When the two atoms are in contact, potassium transfers its outer electron to chlorine which readily accepts it, resulting in both atoms achieving a state of eight outermost electrons. With this electron transfer, the ionic bond in KCl is formed.

Hope this helped!!!

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According to solubility rules, which of the following compounds is soluble in water?
Nata [24]
The answer is A hope this help
6 0
3 years ago
Anything in red is the question
laila [671]

Answer:

  • Question 19: the three are molecular compounds.
  • Question 20: CuSO₄.5H₂O

Explanation:

<em>Question 19.</em>

  • C₂H₄
  • HF
  • H₂O₂

All of them are the combination of two kinds of different atoms in fixed proportions.

  • C₂H₄: two carbon atoms per four hydrogen atoms
  • HF: one hydrogen atom per one fluorine atom
  • H₂O₂: two hydrogen atoms per two oxygent atoms

Thus, they all meet the definition of compund: a pure substance formed by  two or more different elements with a definite composition.

Molecular compounds are formed by covalent bonds and ionic compounds are formed by ionic bonds.

Two non-metal elements, like H-F, C - C, C - H, H-O, H - H, and O - O will share electrons forming covalent bonds to complete their valence shell. Thus, the three compounds are molecular and not ionic.

<em>Question 20. </em>Formula of copper(II) sulfate hydrate with 36.0% water.

Copper(II) sulfate is CuSO₄. Its molar mass is 159.609g/mol

Water is H₂O. Its molar mass is 18.015g/mol

Calling x the number of water molecules in the hydrate, the percentage of water is:

       \dfrac{18.015x}{18.015x+159.609}=36\%\\ \\ \\ \dfrac{18.015x}{18.015x+159.609}=0.36

From which we can solve for x:

      18.015x=6.4854x+57.45924\\ \\ 11.5296x=57.45924\\ \\ x\approx4.98\approx5

Thus, there are 5 molecules of water per each unit of CuSO₄, and the formula is:

  • CuSO₄.5H₂O
4 0
3 years ago
If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho
ankoles [38]

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol  at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we  have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

n = \frac{PV}{RT}

n = \frac{0.078 * 15000}{0.082*290}

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

number of moles = \frac{&85}{46.07}

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

5 0
3 years ago
The following initial rate data are for the reaction of hypochlorite ion with iodide ion in 1M aqueous hydroxide solution: OCI+r
Vinil7 [7]

Answer:

Rate = k [OCl] [I]

Explanation:

OCI+r → or +CI

Experiment [OCI] M I(-M) Rate (M/s)2

1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3

2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3

3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3

4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3

The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.

In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.

In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.

The rate law is given as;

Rate = k [OCl] [I]

5 0
3 years ago
A 20-ohm resistor is the loud connected to a voltage source of 220 volts. what current will flow through the conductor
grandymaker [24]
200 volts of current will flow through the conductor. 
5 0
3 years ago
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