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Based on location in the periodic table, which element has chemical properties that are most similar to iodine

Ymorist [56]

Based on location in the periodic table, fluorine (F) has chemical properties that are most similar to iodine.

In the periodic table, elements are classified in groups and periods. The elements in the same group are chemically similar and they have the same number of valence electrons. Elements in the same period have the same highest energy level.

Hence, when looking for an element that is most chemically similar to iodine, we have to consider the element that is in the same group with iodine.

Both fluorine are iodine are both in group 17 hence, fluorine is most chemically similar to iodine.

Learn more: brainly.com/question/11155928

3 0

3 years ago

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Why are the molecules of hydrocarbons nonpolar?

laiz [17]

<span>This is not the case in the hydrocarbon tail. The electronegativity of hydrogen and carbon are very similar, so the electron cloud is distributed evenly over the two atoms. Carbon-hydrogen bonds are said to be non-polar because they do not have positive and negative poles within themselves. Hope this helps. </span>

4 0

3 years ago

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The powder mixture (Cu, Al and Fe) = 10g was oxidized from sufficient chloride acid. 1) What are the possible reactions to the p

Mandarinka [93]

Answer:

1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

Explanation:

1) Possible reactions

2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Mass of each metal

a) Mass of Cu

The waste was the unreacted copper.

Mass of Cu = 2.5 g

b) Masses of Al and Fe

We have two relations :

Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g

H₂ from Al + H₂ from Fe = 6.38 L at NTP

i) Calculate the moles of H₂

NTP is 20 °C and 1 atm.

$\begin{array}{rcl}pV & = & n RT\\\text{1 atm} \times \text{6.38 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{293.15 K}\\6.38 & = & 24.06n \text{ mol}^{-1} \\n & = & \dfrac{6.38}{24.06 \text{ mol}^{-1} }\\\\ & = & \text{0.2652 mol}\\\end{array}$

(ii) Solve the relationship

Let x = mass of Al. Then

7.5 - x = mass of Fe

Moles of Al = x/27

Moles of Fe = (7.5 - x)/56

Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x /18

Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56

∴ x/18 + (7.5 - x)/56 = 0.2652

56x + 18(7.5 - x) = 267.3

56x + 135 - 18x = 267.3

38x = 132.3

x = 3.5 g

Mass of Al = 3.5 g

Mass of Fe = 7.5 g - 3.5 g = 4.0 g

The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

4 0

3 years ago

By titration, it is found that 31.7 mL of 0.145 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the concentrati

arlik [135]

Answer:

0.184 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

HCl + NaOH —> NaCl + H2O

From the balanced equation above, the following data were obtained:

Mole ratio of the acid, HCl (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Next, the data obtained from the question. This includes:

Volume of the base, NaOH (Vb) = 31.7 mL Molarity of the base, NaOH (Mb) = 0.145 M

Volume of the acid, HCl (Va) = 25.0 mL

Molarity of the acid, HCl (Ma) =?

Finally, we shall determine the molarity of the acid (HCl) as shown below:

MaVa /MbVb = nA/nB

Ma × 25 / 0.145 × 31.7 = 1

Cross multiply

Ma × 25 = 0.145 × 31.7

Ma × 25 = 4.5965

Divide both side by 25

Ma = 4.5965 / 25

Ma = 0.184 M

Therefore, the molarity of the acid (HCl) is 0.184 M

6 0

3 years ago

Which mixture can be classified as a homogeneous mixture?

kramer

Concrete is classified as a homogeneous mix

4 0

3 years ago