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GaryK [48]
3 years ago
13

Is -280 f degrees hot or cold

Chemistry
2 answers:
N76 [4]3 years ago
6 0

Answer:

COLD

Explanation:

Nimfa-mama [501]3 years ago
6 0

Answer:

hot

Explanation:

anything above 100 is consider hot

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NEED HELP!!!!!!
Mkey [24]
Chlorine.

If you search Google images for "Aufbau principle periodic table," you'll find some handy diagrams that will make it much easier to determine an element based on its electron configuration. Determine the number of electrons in the last part of the configuration (in this case, 5), locate that group on the Aufbau periodic table diagram, then count that number from left to right within that group. In this case, within the "3p" portion of the Periodic table, count to 5 and you'll find Chlorine as the answer.
6 0
3 years ago
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Which state of matter has the highest kinetic energy
kvasek [131]

Answer:

gases

Explanation:

Their atoms are the least far apart

4 0
3 years ago
A chemistry teacher adds 50.0 mL of 1.50 M H2SO4 solution to 200 mL of water. What is the concentration of the final solution? U
Aleks [24]
Hello!

The concentration of the final solution when a<span> chemistry teacher adds 50.0 mL of 1.50 M H2SO4 solution to 200 mL of water is 0,3 M

To calculate that, you'll need to use the dilution law, where initial and final concentrations are M1 and M2 respectively, and initial and final volumes are V1 and V2, as shown below. Keep in mind that the final volume is the sum of the 200 mL of water and the 50 mL of H</span>₂SO₄ that were added by the teacher. 

M2= \frac{M1*V1}{V2}= \frac{1,50 (mol H_2SO_4/L)*50mL}{(50 mL + 200 mL)}=0,3(mol H_2SO_4/L)

Have a nice day!
6 0
3 years ago
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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

6 0
3 years ago
Which of these types of changes is a physical change?
Leto [7]

The correct answer is A. Can I get the Brainliest?

6 0
3 years ago
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