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-Dominant- [34]
2 years ago
14

If the specific heat of water is 4. 18 j/g∙°c, how much heat is required to increase the temperature of 1. 20 kg of water from 2

3. 00 °c to 39. 00 °c?
Chemistry
1 answer:
Mkey [24]2 years ago
8 0

The heat required to increase the temperature of 1.20 g of water is  80,256 k/j

<h3>What is specific heat?</h3>

The amount of energy needed to raise the temperature of one gram of a substance by one degree Celsius.

4. 18 J heat is required to change the temperature of 1.20 kg of water from 23. 00 °c to 39. 00 °c.

The heat required to raise temperature is the product of mass, specific heat and temperature change

1,200 × 4.18 × (39 − 23) = 80,256

Thus, the heat required to increase the temperature of 1.20 g of water is  80,256 k/j

Learn more about specific heat

brainly.com/question/11297584

#SPJ4

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Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit
ser-zykov [4K]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density = 10.28g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell =?

Putting values in above equation, we get:

10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm

Conversion factor used:  1cm=10^{10}pm  

Hence, the edge length of the unit cell is 314 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 314}{4}=135.9pm

Hence, the radius of the molybdenum atom is 135.9 pm

4 0
3 years ago
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