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Levart [38]
3 years ago
5

Property by virtue of which metals can be beaten into shapes

Chemistry
1 answer:
Anit [1.1K]3 years ago
6 0

The property which allow the metals to be hammered into thin sheets is called malleability. Gold and silver are the best malleable metals. It is due to the property of malleability that metals can be bent to form objects of different shapes by beating with a hammer.


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Which scenario describes an interaction between two of Earth's spheres?
quester [9]

The correct answer is D. Bears dig big holes in the ground to protect their young

Explanation:

The Earth spheres include the biosphere (life in the Earth), the hydrosphere (water bodies), the geosphere (rocks and other elements that compose land and soil), and the atmosphere (gases that compose the air). In this context, there is an interaction between two spheres: the biosphere and the geosphere, when a bear digs holes in the ground because a living organism that is part of the biosphere is modifying the structure and shape of superficial soil, which is part of the geosphere.

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2 years ago
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Which element has the lowest first ionization energy?<br> Be<br> Sr<br> OCa<br> Mg
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Answer
I think Sr
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3 years ago
Kinetic energy is the energy of motion.<br><br> a) True<br><br> b) False
FinnZ [79.3K]
The answer is a) true 
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3 years ago
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A gas occupies 800ml at a temperature of 27C. What is the volume at 132C?
pishuonlain [190]

V
1
​
/T
1
​
=V
2
​
/T
2
​

(900.0 mL) / (300.0 K) = (x) / (405.0 K); x = 1215 mL.

Change the 900 to 800, and the 300 to 27, then change the 405 to 132. And solve
3 0
3 years ago
Calculate the change in energy when 75.0 grams of water drops from<br> 31.0C to 21.6.
zysi [14]

Answer: Step 1: Calculate qsur (the surrounding is

usually the water)

qsur = ? J

m = 75.0 g water

c = 4.184 J/g

oC

ΔT = (Tfinal- Tinitial)= (21.6 – 31.0) = -9.4 oC

qsur = m · c · (ΔT)

qsur = (75.0g) (4.184 J/g

oC) (-9.4 oC)

qsur = - 2949.72 J

First, using the information we know that we

must solve for qsur, which is the water. We know

the mass for water, 75.0g, the specific heat of

the water, 4.184 j/g

o

c, and the change in

temperature, 21.6-31.0 = -9.4 oC. Plugging it

into the equation, we solve for qsur.

Step 2: Calculate qsys qsys = - (qsur)

qsys = - (- 2949.72 J)

qsys = + 2949.72

In this case, the qsur is negative, which means

that the water lost energy. Where did it go? It

went to the system. Thus, the energy of the

system is negative, opposite, the energy of the

surrounding.

Step 3: Calculate moles of the substance

that is the system

Given: 12.8 g KCl

Mol system = (g system given)

(molar mass of system)

Mol system = (12.8 g KCl)

(39.10g + 35.45g)

Mol system = 12.8 g KCl

74.55 g

Mol system = 0.172

Here, we solve for the mol in the system by

using the molar mass of the material in the

system.

Step 4: Calculate ΔH ΔH = q sys .

Mol system

ΔH= + 2949.72 J

0.172 mol

ΔH= +17179.81 J/mol or +1.72 x 104

J/mol

i hope this helps

5 0
3 years ago
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