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Sav [38]
4 years ago
13

Programs that award certificate that u can help in securing jobs are called​

Chemistry
1 answer:
leonid [27]4 years ago
7 0

vocational or technical education programs.

Explanation:

A vocational or technical education program is one that awards certificate that can help in securing jobs.

In a vocational or technical program, one is studying for a specific job role. It involves and hands on practical processes within a short period of time possible that makes one viable for some roles.

Examples are IT technician, Welders, Carpentry etc

learn more:

Job hazards brainly.com/question/3193019

#learnwithBrainly

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Electrons are deflected by a magnetic field because...
max2010maxim [7]

they have a negligible mass and they also have a negative charge so they have an association with the magnetic field. Do you remember polar? So electrons are negative and they attract the nucleus which is overall postivie (since protons are positive) so they act as magnetic since they both have an attraction

5 0
3 years ago
When drawing a Bohr model for Sulfur, how many energy levels will you
grandymaker [24]

Answer:

3

Explanation:

2,8,6

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3 years ago
Can someone solve this problem 5
Westkost [7]

Answer:

2

Step-by-step explanation:

A. Moles before mixing

<em>Beaker I: </em>

Moles of H⁺ = 0.100 L × 0.03 mol/1 L

                   = 3 × 10⁻³ mol

<em>Beaker II: </em>

Beaker II is basic, because [H⁺] < 10⁻⁷ mol·L⁻¹.

        H⁺][OH⁻] = 1 × 10⁻¹⁴   Divide each side by [H⁺]

             [OH⁻] = (1 × 10⁻¹⁴)/[H⁺]

             [OH⁻] = (1 × 10⁻¹⁴)/(1 × 10⁻¹²)

             [OH⁻] = 0.01 mol·L⁻¹

Moles of OH⁻ = 0.100 L × 0.01 mol/1 L

                      = 1 × 10⁻³ mol

B. Moles after mixing

                 H⁺    +    OH⁻   ⟶ H₂O

I/mol:      3 × 10⁻³   1 × 10⁻³

C/mol:   -1 × 10⁻³  -1 × 10⁻³

E/mol:    2 × 10⁻³          0

You have more moles of acid than base, so the base will be completely neutralized when you mix the solutions.

You will end up with 2 × 10⁻³ mol of H⁺ in 200 mL of solution.


C. pH

 [H⁺] = (2 × 10⁻³ mol)/(0.200 L)

        = 1 × 10⁻² mol·L⁻¹

 pH = -log[H⁺ ]

       = -log(1 × 10⁻²)

       = 2

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