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ikadub [295]
3 years ago
11

Which is the BEST evidence that a chemical change has occurred?

Chemistry
1 answer:
MaRussiya [10]3 years ago
3 0

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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
A student adds a rectangular block of mass 26.10 g to a graduated cylinder initially filled with 40.1 mL of water, but some wate
Amiraneli [1.4K]

Answer:

The splash cause the volume of the block to be too high

Explanation:

This is because, from the previous measurement made by the student, the volume of the block is 5.7ml higher.

8 0
2 years ago
A student collected 7.46 moles of I2 gas at STP How many liters is that?
sukhopar [10]

Answer:

for more details are in the pic

6 0
3 years ago
Which element will most easily lose an electron? A. Calcium (Ca) B. Potassium (K) C. Boron (B) D. Krypton (Kr)
Artemon [7]
<h2>Answer:</h2>

Option B. Potassium(K).

<h2>Explanation:</h2>

Electronic configuration of the given elements are:

  1. Ca - [Ar] 4s²
  2. K - [Ar] 4s¹
  3. B - [He] 2s2 2p1
  4. Kr -  [Ar] 3d¹⁰4s²4p⁶
  1. Krypton(K) have 36 electrons and it is a noble gas and hence all of its shells are completely filled with electrons and hence it will never loose electrons in normal conditions.
  2. Boron(B) have 5 electrons and 3 electrons in its outer shell. In order to attain a stable configuration it will loose 3 electrons and it is difficult to loose 3 electrons at a time for an atom.
  3. Calcium(Ca) have 20 electrons and 2 electrons in its outermost shell, in order to attain a stable configuration it will loose 2 electrons. it is quite difficult but easier than Boron.
  4. Potassium(K) have 21 electrons and 1 electron in its outermost orbit and in order to attain a stable configuration it will loose 1 electron. It is much easier to donate 1 electron than 2 or 3 electrons.

Result: Potassium will loose an electron most easily from the given elements.

8 0
3 years ago
Read 2 more answers
For a reaction A + B → products, the following data were collected. Experiment Number Initial Concentration of A (M) Initial Con
Afina-wow [57]

Answer:

Rate constant k = 1.57*10⁻⁵ s⁻¹

Explanation:

Given reaction:

A\rightarrow B

Expt    [A] M        [B] M         Rate [M/s]

1          3.40         4.16           1.82*10^-4

2         4.59         4.16           3.32*10^-4

3.        3.40          5.46          1.82*10^-4

Rate = k[A]^{x}[B]^{y}

where k = rate constant

x and y are the orders wrt to A and B

To find x:

Divide rate of expt 2 by expt 1

\frac{3.32*10^{-4} }{1.82*10^{-4} } =\frac{[4.59]^{x} [4.16]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\x =2

To find y:

Divide rate of expt 3 by expt 1

\frac{1.82*10^{-4} }{1.82*10^{-4} } =\frac{[3.40]^{x} [5.46]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\y =0

Therefore: x = 2, y = 0

Rate = k[A]^{2}[B]^{0}

To find k

Use rate for expt 1:

k = \frac{Rate1}{[A]^{2} } =\frac{1.82*10^{-4}M/s }{[3.40]^{2} } =1.57*10^{-5} s-1

6 0
3 years ago
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