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nordsb [41]
3 years ago
5

Which element is in the same family as Chlorine (Cl) and Fluorine (F)?

Chemistry
2 answers:
ValentinkaMS [17]3 years ago
8 0

Answer/Explanation:

Chlorine and Fluorine are in the Halogen family. The elements in the Halogen family are:

Fluorine (F)

Chlorine (Cl)

Bromine (Br)

Iodine (I)

Astatine (At)

Tennessine (Ts)

Hydrogen (H) is a nonmetal

Oxygen (O) is a nonmetal

Lithium (Li) is an alkaline metal.

DedPeter [7]3 years ago
5 0

Answer/Explanation:

Chlorine and Fluorine are in the Halogen family. The elements in the Halogen family are:

Fluorine (F)

Chlorine (Cl)

Bromine (Br)

Iodine (I)

Astatine (At)

Tennessine (Ts)

Hydrogen (H) is a nonmetal

Oxygen (O) is a nonmetal

Lithium (Li) is an alkaline metal.

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Find the mass of 0.456 mol of Ca(OH)2 <br> Please if you can show the work.
m_a_m_a [10]

Answer:

Ca(OH)2 molecular weight. Molar mass of Ca(OH)2 = 74.09268 g/mol. This compound is also known as Calcium Hydroxide. Convert grams Ca(OH)2 to moles or moles Ca(OH)2 to grams. Molecular weight calculation: 40.078 + (15.9994 + 1.00794)*2 ››

3 0
2 years ago
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What is the pH of a solution with an [H+] of (a) 5.4 x 10-10, (b) 4.3 x 10-5, (c) 5.4 x 10-7?
IgorLugansk [536]

Answer:

a. 9.2

b. 4.4

c. 6.3

Explanation:

In order to calculate the pH of each solution, we will use the definition of pH.

pH = -log [H⁺]

(a) [H⁺] = 5.4 × 10⁻¹⁰ M

pH = -log [H⁺] = -log 5.4 × 10⁻¹⁰ = 9.2

Since pH > 7, the solution is basic.

(b) [H⁺] = 4.3 × 10⁻⁵ M

pH = -log [H⁺] = -log 4.3 × 10⁻⁵ = 4.4

Since pH < 7, the solution is acid.

(c) [H⁺] = 5.4 × 10⁻⁷ M

pH = -log [H⁺] = -log 5.4 × 10⁻⁷ = 6.3

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2 years ago
How to put Roman numerals in a science equation
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Example:

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A chemist is making 200 l of a solution that is 62% acid. he is mixing an 80% acid solution with a 30% acid solution. how much o
Savatey [412]
Answer: 72L of 30% and 128L of 80%

You can determine the weight of the acid by multiplying the concentration with the volume. Let say v1 is the volume of 30% solution needed and v2 is the volume of 80% solution.
The weight of acid from the used solution should be equal to the product. You can get this equation
final solution= solution1 + solution2
200l * 62%= v1 * 30% + v2*80%
124L= 0.3v1 + 0.8v2
124L- 0.3v1= 0.8v2
v2=155L- 0.375v1

The total volume of both should be 200l. If you use the previous equation, you can calculate:
v1+v2=200L
v1+ (155L- 0.375v1)= 200L
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v1+v2=200L
v2= 200L- 72L= 128L
5 0
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