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Semenov [28]
4 years ago
15

What is the practical use of scientific knowledge to solve everyday problems and improve lives?

Chemistry
1 answer:
Aleks04 [339]4 years ago
5 0
A. Research is the answer
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how did the gross colors (continuous spectrum imparted to the flame by the metal ion solutions compare with the individual spect
Anna35 [415]

The gross colors can be imparted to the flame by the metal ion solutions .

1.To serve as an excitation source, turn on a Bunsen burner.

2. Perform flame tests on the chloride solutions of Li+, Na+, K+, Ca2+, Sr2+, and Ba2+ as instructed by your lab instructor. Wearing the unique glass blower glasses, take note of the obscene hue that each ion contributes to the flame. With one exception: in order to view the sodium flame, you must remove the special glasses that suppress sodium emissions.

3. Obtain two unidentified answers and note their numerical values. Using the ugly hue the solution gives the flame as a guide, identify the metal ions that are present.

One of the six ions you tested will make up the particle.

To know more about continuous spectrum, please refer:

brainly.com/question/28032005

#SPJ4

7 0
1 year ago
Which of the following is a direct result of hydrogen bonding?
prisoha [69]
Cohesion and adhesion
5 0
3 years ago
Give the volume indicated. Be sure to include all significant figures.<br> 9.2<br> 9.16<br> 9.26
Gemiola [76]

Answer:

The volume indicated would be closest to 9.16

3 0
3 years ago
Read 2 more answers
Calculate the Molarity (M, moles solute / Liter of solution) of a solution that is made by dissolving 3.456 mmoles of Barium per
yan [13]
Since Molarity is moles per Liter, you will first need to convert the mL into L, which makes it 0.4321L. Next divide: moles on top, Liters on the bottom.
3.456 moles / .4321L = 7.998 M
3 0
4 years ago
(a)Determine the number of KNO3 molecules in 0.750 mol KNO3.
svp [43]

A. The number of molecules in 0.750 mole of KNO₃ is 4.515×10²³ molecules

B. The mass (in milligrams) of 2.39×10²⁰ molecules of Ag₂SO₄ is 124 mg

C. The number of molecules in 3.429 g of NaHCO₂ is 3.04×10²² molecules

<h3>Avogadro's hypothesis </h3>

1 mole of substance = 6.02×10²³ molecules

<h3>A. How to determine the number of molecules </h3>

1 mole of KNO₃ = 6.02×10²³ molecules

Therefore,

0.750 mole of KNO₃ = 0.75 × 6.02×10²³

0.750 mole of KNO₃ = 4.515×10²³ molecules

<h3>B. How to determine the mass of Ag₂SO₄</h3>

6.02×10²³ molecules = 312 g of Ag₂SO₄

Therefore,

2.39×10²⁰ molecules = (2.39×10²⁰ × 312) / 6.02×10²³

2.39×10²⁰ molecules = 0.124 g

Multiply by 1000 to express in mg

2.39×10²⁰ molecules = 0.124 g × 1000

2.39×10²⁰ molecules = 124 mg of Ag₂SO₄

<h3>C. How to determine the number of molecules </h3>

68 g of NaHCO₂ = 6.02×10²³ molecules

Therefore,

3.429 g of NaHCO₂ = (3.429 × 6.02×10²³) / 68

3.429 g of NaHCO₂ = 3.04×10²² molecules

Learn more about Avogadro's number:

brainly.com/question/26141731

5 0
2 years ago
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