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ira [324]
3 years ago
12

ELEMENTS ARE CLASSIFIED AS METALS, NONMETALS, AND METALLOIDS. USE THE PERIODIC TABLE AND IDENTIFY TWO OF EACH CLASSIFICATION.

Chemistry
1 answer:
alekssr [168]3 years ago
3 0

Answer:

Elements are classified into- metals – Nonmetals- Metalloids – Noble gases. State which of A, B, C, D is a:

1) Metallic element

2) Non-metallic element

3) Metalloid

4) Noble gas.

A) Is non-malleable, non-ductile and a poor conductor of electricity

B) Has lustre, is malleable and ductile and a good conductor of electricity

C) Is unreactive and inert and present in traces in air

D) Shows properties of both metals and nonmetals

Explanation:

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If I have 7.5 L of a gas in a piston at a pressure of 1.75 atm and compress the gas until its volume is 3.7 L, what will the new
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Answer:

3.55atm

Explanation:

We will apply Boyle's law formula in solving this problem.

P1V1 = P2V2

And with values given in the question

P1=initial pressure of gas = 1.75atm

V1=initial volume of gas =7.5L

P2=final pressure of gas inside new piston in atm

V2=final volume of gas = 3.7L

We need to find the final pressure

From the equation, P1V1 = P2V2,

We make P2 subject

P2 = (P1V1) / V2

P2 = (1.75×7.5)/3.7

P2=3.55atm

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A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4
Stella [2.4K]

Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron =  560 J/(kg.K)

c_1 = specific heat of water = 4186 J/(kg.K)

m_1 = mass of iron = 825 g

m_2 = mass of water = 40 g

T_f = final temperature of water and iron = ?

T_1 = initial temperature of iron = 352^oC=273+352=625K

T_2 = initial temperature of water = 20^oC=273+20=293K

Now put all the given values in the above formula, we get:

(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)

T_f=537.12K

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

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