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3 years ago

Define Triple point of water?

2 answers:

6 0

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2><h2>TRIPLE POINT:</h2>

Triple point is that temperature, at which the substance is found in all three states i.e. solid, liquid and gas. Every substance has different triple point, because Triple point is just above the freezing point and each substance has different freezing and boiling point.

<h2>_____________________________________</h2><h2>WATER:</h2>

Water has a Melting point or Freezing Point (MP or FP) of $0^0$ C or 273K and Boiling point (BP) $100^{0}$ C or 373 K. Since the triple point is just above the MP or FP thus Triple Point of water is $0.01^{0}$ C or 273.16K.

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2><h2 />

tigry1 [53]3 years ago

6 0

Answer:

0.01 degrees centigerade

Explanation:

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A sample of an unknown gas takes 222 s to diffuse through a porous plug at a given temperature. At the same temperature, N2(g) t

gulaghasi [49]

Answer:

$\large \boxed{\text{45.1 g/mol}}$

Explanation:

Graham’s Law applies to the diffusion of gases:

The rate of diffusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

$r \propto \dfrac{1}{\sqrt{M}}$

If you have two gases, the ratio of their rates of diffusion is

$\dfrac{r_{2}}{r_{1}} = \sqrt{\dfrac{M_{1}}{M_{2}}}$

The time for diffusion is inversely proportional to the rate.

$\dfrac{t_{2}}{t_{1}} = \sqrt{\dfrac{M_{2}}{M_{1}}}$

Data:

t₂ = 222 s

t₁ = 175 s

M₁ = 28.01

Calculation :

$\begin{array}{rcl}\dfrac{222}{175} & = & \sqrt{\dfrac{M_{2}}{28.01}}\\\\1.269 & = & \sqrt{\dfrac{M_{2}}{28.01}}\\\\1.609 & = & \dfrac{M_{2}}{28.01}\\\\M_{2} & = & 1.609 \times 28.01\\ & = & \textbf{45.1 g/mol}\\\end{array}\\\text{The molar mass of the unknown gas is $\large \boxed{\textbf{45.1 g/mol}}$}$

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3 years ago

A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 4250 mL of water at 2

Assoli18 [71]

Answer:

The specific heat of copper is 0.385 J/g°C

Explanation:

A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 425 mL of water at 22.55 degrees Celsius. The final temperature of the water was recorded to be 26.15 degrees Celsius. What is the specific heat of the copper?

Step 1: Data given

Mass of copper = 85.2 grams

Temperature of copper = 221.32 °C

Volume of water = 425 mL

Temperature of water = 22.55 °C

Final temperature = 26.15 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculat the specific heat of copper

Heat lost = heat gained

Q = m*c*ΔT

Qcopper = -Qwater

m(copper)*c(copper)*ΔT(copper) = - m(water) * c(water) * ΔT(water)

⇒ m(copper) = 85.2 grams

⇒ c(copper) = TO BE DETERMINED

⇒ ΔT(copper) = the change in temeprature = T2 -T1 = 26.15 -221.32 = -195.17 °C

⇒ m(water) = The mass of water = 425 mL * 1g/mL = 425 grams

⇒ c(water) = The specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature of water = 26.15 - 22.55 = 3.6

85.2 * c(copper) * (-195.17) = -425 * 4.184 * 3.6

c(copper) = 0.385 J/g°C

The specific heat of copper is 0.385 J/g°C

(Note, The original question says the volume of the water is 4250 mL. IF this is not an error, the specific heat of copper is 3.85 J/g°C (10x higher than the normal value).

8 0

3 years ago

Define Triple point of water?​

Define Triple point of water?