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Alika [10]
3 years ago
14

NEED HELP ASAP WILL GIVE BRAINLIEST. What is the element that has the atomic number of 17?

Chemistry
1 answer:
salantis [7]3 years ago
5 0

Hey buddy the answer is Chlorine ..

plzz follow me I'll help u more if u need

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Can somebody plz help answer these true or false questions correctly thanks! (Only if u for sure know them) :)
asambeis [7]

Answer:

1. False

2. True

3. False

4. True

5. True

Explanation:

7 0
3 years ago
Consider the reaction given below.
Drupady [299]

Answer:

  • <u>K =  0.167 s⁻¹</u>

Explanation:

<u>1) Rate law, at a given temperature:</u>

  • Since all the data are obtained at the same temperature, the equilibrium constant is the same.

  • Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

        r = K [A]ᵃ [B]ᵇ

<u>2) Use the data from the table</u>

  • Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

         Divide r₂ by r₁:     [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

  • Use the first and second set of data to find the exponent a:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

        Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

                                  2ᵃ = 2 ⇒ a = 1

         

<u>3) Write the rate law</u>

  • r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

<u>4) Use any set of data to find K</u>

With the first set of data

  • r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K =  0.167 s⁻¹

6 0
3 years ago
Which is an example of potential energy?
Setler79 [48]
Probably C hope this helps
5 0
2 years ago
Read 2 more answers
Describe the process you used to build a model. What did you do first? second?
dimaraw [331]
First read the introduction.

Seconds look at the pictures how to build it.
3 0
3 years ago
How much energy (in J) is lost when a sample of iron with a mass of 26.4 g cools from 74.0 ∘C to 26.0 ∘C?
Greeley [361]

Answer:

Q=-526.6J

Explanation:

Hello,

In this case, for the computation of the energy loss when the cooling process is carried out, we use the shown below equation:

Q=mCp\Delta T

Whereas we need the mass, specific heat and change in temperature of iron within the process. Thus, the only value we need is the specific heat that is 0.444 J/(g°C), therefore, we compute the heat loss:

Q=26.4g*0.444\frac{J}{g\°C}*(26.0\°C-74.0\°C)\\ \\Q=-526.6J

Negative sign points out the loss due to the cooling.

Regards.

8 0
3 years ago
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