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Anon25 [30]
3 years ago
5

Write the orbital notation for the following elements. a. P b. B c. Na d. C

Chemistry
2 answers:
Zigmanuir [339]3 years ago
8 0
C maybe is the answer
Veseljchak [2.6K]3 years ago
4 0
I think that the answer is C
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How many moles of solute are in 300 mL of 1.5 M CaCl2? How many grams fo CaCl2 is this?
tester [92]

Answer:

1. 0.45 mole

2. 49.95g

Explanation:

The following were obtained from the question:

Volume of solution = 300mL = 300/1000 = 0.3L

Molarity = 1.5 M

Mole of CaCl2 =?

1. We can obtain the mole of the solute as follow:

Molarity = mole of solute /Volume of solution

1.5 = mole of solute/0.3

Mole of solute = 1.5 x 0.3

Mole of solute = 0.45 mole

2. The grams in 0.45 mole of CaCl2 can be obtained as follow:

Molar Mass of CaCl2 = 40 + (35.5 x 2) = 40 + 71 = 111g/mol

Mole of CaCl2 = 0.45 mole

Mass of CaCl2 =?

Mass = number of mole x molar Mass

Mass of CaCl2 = 0.45 x 111

Mass of CaCl2 = 49.95g

3 0
3 years ago
Read 2 more answers
Morphine has the formula c17h19no3. it is a base and accepts one proton per molecule. it is isolated from opium. a 0.685 −g samp
kotykmax [81]
2 C₁₇H₁₉NO₃ + H₂SO₄ → Product
Moles of H₂SO₄ = M x V(liters) = 0.0116 x 8.91/1000 = 1.033 x 10⁻⁴ mole 
moles of morphine = 2 x moles of H₂SO₄ = 2.066 x 10⁻⁴
Mass of morphine = moles x molar mass of morphine = 2.066 x 10⁻⁴ x 285.34 
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percent morphine = \frac{mass of morphine}{mass of opium} x 100
                             = \frac{0.059}{0.685} x 100 = 8.6 %   
7 0
3 years ago
When 20 ml of 0.1 M HCl is mixed with 20
Bas_tet [7]

Answer:

it is b i took the test

Explanation:

4 0
2 years ago
What is the abbreviation for the element with atomic number 11?
VikaD [51]

Answer:

Sodium (Na) has atomic number 11.

5 0
3 years ago
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How many liters of fluorine gas, at standard temperature and pressure, will react with 23.5 grams of potassium metal? Show all o
natka813 [3]
The balanced chemical reaction is written as:

2K + F2 = 2KF

We are given the amount of potassium to be used in the reaction. THis will be the starting point. We do as follows:

23.5 g K ( 1 mol / 39.10 g) (1 mol F2 / 2 mol K ) ( 22.4 L / 1 mol ) = 6.73 L F2 gas needed
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3 years ago
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