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user100 [1]
4 years ago
6

Who discovered the flame test

Chemistry
1 answer:
Genrish500 [490]4 years ago
6 0

Answer:

Thomas Melvill

Explanation:

...

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One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0
3 years ago
Which atom, O or N, has the greater atomic radius?
alukav5142 [94]

Answer:

Nitrogen because it has a greater value of PPM, so the bigger the radii

Explanation:

plz give Brainlyest

8 0
3 years ago
A student collects 25 mL of gas at 0.98 bar. What volume would this gas occupy at 1.013 bar? * 3 points 25.8 mL 24.2 mL .09 mL 3
trasher [3.6K]

Answer:

24.2 mL.

Explanation:

<em>Assuming constant temperature</em>, we can solve this problem using <em>Boyle's law</em>, which states:

  • P₁V₁=P₂V₂

Where:

  • P₁ = 0.98 bar
  • V₁ = 25 mL
  • P₂ = 1.013 bar
  • V₂ = ?

We <u>input the data</u>:

  • 0.98 bar * 25 mL = 1.013 bar * V₂

And <u>solve for V₂</u>:

  • V₂ = 24.18 mL

The closest option is the second one: 24.2 mL.

3 0
3 years ago
The density of pure gold (Au) is 19.3 g/cm3. What is the mass of a block of gold that measures 1.2 cm x 4.00 cm x 3.20 cm? Pleas
N76 [4]

Answer:

296.448 g

Explanation:

Density:

we know that density of an object can be measured by dividing its mass to its volume.

Symbol:

The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m= mass

V= volume

Given data:

volume= 1.2 cm x 4.00 cm x 3.20 cm =15.36 cm3

density=  19.3 g/cm3

mass=?

Now we will put the values in the formula,

d=m/v

d x V= m

m = 15.36 cm3 x 19.3 g/cm3

m=296.448 g

so the mass of block of gold is 296.448 g

8 0
3 years ago
Ba(CN) _<br> What is the compound
AleksandrR [38]
Barium Cynanide is the compound
7 0
4 years ago
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