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3 years ago

What is the variable for the quantity of heat in a system?

Chemistry

1 answer:

Crank3 years ago

8 0

Answer:

The heat capacity is the amount of heat, expressed usually in Joules or calories, needed to change the system by 1 degree Celsius. The specific heat capacity is the amount of heat required to raise 1 gram of a substance by 1 degree Celsius.

Explanation:

For example, the specific heat of H2O(l) is 4.18 J/g ° C.

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Three blocks are shown: Three blocks are shown. Block A has mass 3 kilograms, length 6 centimeters, height 4 centimeters, and wi

dsp73

Answer:

Block A has the greatest density.

Explanation:

Three blocks are shown: Three blocks are shown.

Block A has mass 3 kilograms, length 6 centimeters, height 4 centimeters, and width 2 centimeters.

Block B has mass 1 kilograms, length 6 centimeters, height 2 centimeters, and width 4 centimeters.

Block C has mass 2 kilograms, length 2 centimeters, height 6 centimeters, and width 4 centimeters.

Which statement is correct?

- Block A has the greatest density.

- Block B has the greatest density.

- The density of Block A is equal to the density of Block B.

- The density of Block B is equal to the density of Block C

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1 year ago

In a chemical reaction, which of the following is true of the products?

zalisa [80]

Your answer would be c I’m pretty sure

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2 years ago

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What is a cool 1 week or less science fair project for grade 9?

Leno4ka [110]

Make a fizzing volcano with facts and pictures of a volcano

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3 years ago

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The volume density of atoms for a bcc lattice is 5 x 1026 m-3. Assume that the atoms are hard spheres with each atom touching it

ollegr [7]

Explanation:

It is known that for a body centered cubic unit cell there are 2 atoms per unit cell.

This means that volume occupied by 2 atoms is equal to volume of the unit cell.

So, according to the volume density

$5 \times 10^{26} atoms = 1 [tex]m^{3}$

2 atoms = $\frac{1 m^{3}}{5 \times 10^{26} atoms} \times 2 atoms$

= $4 \times 10^{-27} m^{3}$

Formula for volume of a cube is $a^{3}$ . Therefore,

Volume of the cube = $4 \times 10^{-27} m^{3}$

As lattice constant (a) = $(4 \times 10^{-27} m^{3})^{\frac{1}{3}}$

= $1.59 \times 10^{-9} m$

Therefore, the value of lattice constant is $1.59 \times 10^{-9} m$ .

And, for bcc unit cell the value of radius is as follows.

r = $\frac{\sqrt{3}}{4}a$

Hence, effective radius of the atom is calculated as follows.

r = $\frac{\sqrt{3}}{4}a$

= $\frac{\sqrt{3}}{4} \times 1.59 \times 10^{-9} m$

= $6.9 \times 10^{-10} m$

Hence, the value of effective radius of the atom is $6.9 \times 10^{-10} m$ .

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3 years ago

I need the answers to my chemistry

Rudiy27

Answer:

whats ur question

Explanation:

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3 years ago

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