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MissTica
2 years ago
11

Refer to the image please!

Chemistry
1 answer:
Ludmilka [50]2 years ago
8 0

Answer:

9. Electron Affinity

10. The second option....

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Where can fire might be used as a management tool ?
lions [1.4K]
Natural vegetation!!

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3 years ago
24. All elements found on the left side of the Periodic Table of
SashulF [63]

Answer:

Metals are located on the left side of the periodic table and are generally shiny, malleable, ductile, and good conductors.

Explanation:

5 0
3 years ago
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How many molecules of Ca are found in a sample with 0.2 mols?
DochEvi [55]

Answer:

\boxed{1.2 \times 10^{23}\text{ atoms}}

Explanation:

6.023 × 10²³ atoms of Ca are in 1 mol of Ca

\text{No. of atoms} = \text{0.2 mol} \times \dfrac{6.023 \times 10^{23}\text{atoms }}{\text{1 mol }} = \mathbf{1.2 \times 10^{23}} \textbf{ atoms}}\\\\\text{There are }\boxed{\mathbf{1.2 \times 10^{23}} \textbf{ atoms}} \text{ atoms in 0.20 mol of Ca}

6 0
3 years ago
Put the following names in the correct alphabetic indexing order:(1) Topper & Casey Plumbing(2) KST Enterprises(3) Leland an
dangina [55]

Answer: the correct option is 2, 3, 4, 1.

Explanation:alphabetic indexing order is the order in which files or names are being arranged according to the alphabet. In the following names:

2) KST Enterprises

3)Leland and Son Graphics

4)Lucinda Topper

1) Topper & Casey Plumbing.

While arranging alphabetically, the first letters are usually considered but in a scenario where alphabet occurs twice( 3 And 4) the second letter is considered. I hope this helps, thanks

5 0
3 years ago
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Determine the final temperature of sample with a specific heat of 1.1 J/g°C and a mass of 385 g if it starts out at a temperatur
Assoli18 [71]

Answer:

T2 =21.52°C

Explanation:

Given data:

Specific heat capacity of sample = 1.1 J/g.°C

Mass of sample = 385 g

Initial temperature = 19.5°C

Heat absorbed = 885 J

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )

885 J = 423.5 J/°C× (T2 - 19.5°C )

885 J / 423.5 J/°C = (T2 - 19.5°C )

2.02°C = (T2 - 19.5°C )

T2 = 2.02°C + 19.5°C

T2 =21.52°C

8 0
3 years ago
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