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WITCHER [35]
2 years ago
6

Put the following elements into five pairs of elements that have similar chemical reactivity: F, Sr, P, Ca, O, Br, Rb, Sb, Li, S

Chemistry
1 answer:
kirill115 [55]2 years ago
7 0

Explanation:

Given elements:

  F, Sr, P, Ca, O, Br, Rb, Sb, Li, S

Elements with the same chemical reactivity will belong to the same group on the periodic table. This implies that elements in the same column will have the same reactivity;

  Li and Rb are both alkali metals in group 1

  Ca and Sr are both alkali earth metals in group 2

  F and Br are halogens in group 7

  O and S are group 6 elements

 P and Sb are both in group 5 on the periodic table

So these groupings show elements with the same chemical properties.

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How many atoms are in KOH?
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Answer:

KOH contains only one K, so a mole of KOH contains one mole of K. How many atoms of K are in 1 mole of KOH? There are 6.022 × 1023 atoms of potassium in every mole of potassium. Since one mole of KOH contains one mole of K, the answer is 6.022×1023 atoms of K.

Explanation:

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3 years ago
Be sure to answer all parts.
MrRissso [65]

Answer: The molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

Explanation:

Molarity is the number of moles of a substance present in liter of a solution.

And, moles is the mass of a substance divided by its molar mass.

(a) Moles of ethanol (molar mass = 46 g/mol) is as follows.

Moles = \frac{mass}{molar mass}\\= \frac{28.5 g}{46 g/mol}\\= 0.619 mol

Now, molarity of ethanol solution is as follows.

Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.619 mol}{4.50 \times 10^{2} \times 10^{-3}L}\\= 1.38 M

(b) Moles of sucrose (molar mass = 342.3 g/mol) is as follows.

Moles = \frac{mass}{molar mass}\\= \frac{21.6 g}{342.3 g/mol}\\= 0.063 mol

Now, molarity of sucrose solution is as follows.

Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.063 mol}{0.067 L}  (1 mL = 0.001 L)\\= 0.94 M

(c) Moles of sodium chloride (molar mass = 58.44 g/mol) are as follows.

Moles = \frac{mass}{molar mass}\\= \frac{6.65 g}{58.44 g/mol}\\= 0.114 mol

Now, molarity of sodium chloride solution is as follows.

Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.114 mol}{0.0962 L}\\= 1.182 M

Thus, we can conclude that the molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

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