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2 years ago

The state of water is determined solely by the conditions of _____.

Chemistry

2 answers:

Arlecino [84]2 years ago

7 0

Answer:

The state of water is determined solely by the conditions of pressure and temperature.

Explanation:

Effect of Temperature:

Temperature has direct effect on the kinetic energy of water molecules. Greater the Temperature greater is the K.E of molecules and lesser is the intermolecular forces between molecules. Therefore, at temperature below 0 °C water exists in solid state, at temperature in between 0 and 100 °C it exists in liquid state and (at sea level) and above 100 °C it exists in vapor state.

Explanation:

Elenna [48]2 years ago

6 0

<h3>Answer:</h3>

The state of water is determined solely by the conditions of <u>pressure and temperature</u>.

<h3>Explanation:</h3>

Effect of Temperature:

Effect of Pressure:

The state of water is also influenced by the external pressure. For example, water can be converted into gas phase by reducing the external pressure (this technique is used in vacuum distillation where solvents are evaporated at temperatures lower than their boiling points) and also, the water can remain in liquid state even above 100 °C if the external pressure is increases (this method is being utilized in daily use pressure cookers).

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2 years ago

x g of metal (equivalent wt=12) was completely dissolved in 100 cc of N/2 HCl. The volume was then made upto 500 cc.25 cc of thi

ad-work [718]

Answer:

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Respirometers were set up that contained yeast and 2% glucose solution. Each respirometer was placed at a different temperature.

Kitty [74]

Temperature is the independent variable in this experiment

Explanation:

A respirometer is a tool used to calculate the rate of breathing of a living organism by calculating the amount of exchange of oxygen and/or carbon dioxide.

When the yeast cells are given sugar in an aerobic (oxygen-containing) environment, then the yeast.

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2 years ago

Calculate the oxidation number of s in S2O8^2-

mixer [17]

Given problem:

S₂O₈²⁻

Find the oxidation number of S;

Oxidation number presents the extent of oxidation of each atom of elements a molecular formular or formula unit or an ionic radical.

For radicals:

"the algebraic sum of all the oxidation numbers of all atoms in an ion containing more than one kind of atom is equal to the charge on the ion "

S₂O₈²⁻; oxidation number of O is usually -2

2(S) + 8(-2) = -2

2S - 16 = -2

2S = -2 + 16

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The oxidation state of S in the radical is +7

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2 years ago

A 2.50-l flask contains a mixture of methane (ch4) and propane (c3h8) at a pressure of 1.45 atm and 20°c. when this gas mixture

Cerrena [4.2K]

Answer:- Mole fraction of methane in the original gas mixture is 0.854.

Solution:- From given volume, pressure and temperature, we could calculate the total moles of the gaseous mixture of methane and propane using ideal gas law as:

PV = nRT

$n=\frac{PV}{RT}$

V = 2.50 L

P = 1.45 atm

T = 20 + 273 = 293 K

Let's plug in the values in the equation:

$n=(\frac{1.45*2.50}{0.0821*293})$

n = 0.151

Let's say the solution has X moles of methane. Then moles of propane would be = (0.151 - X)

The combustion equations of methane and propane are:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

From methane balanced equation, there is 1:1 mol ratio between methane and carbon dioxide. So, X moles of methane would produce X moles of carbon dioxide.

From balanced equation of propane, there is 1:3 mol ratio between propane and carbon dioxide. So, (0.151 - X) moles of propane would give 3(0.151 - X) moles of carbon dioxide.

So, total moles of carbon dioxide that we would get from methane and propane combustion are:

X + 3(0.151 - X)

From given data, 8.60 g of carbon dioxide are formed by the combustion of gas mixture.

moles of Carbon dioxide = $8.60g(\frac{1mol}{44g})$

moles of carbon dioxide = 0.195 mol

Hence, 0.195 = X + 3(0.151 - X)

Let's solve this for X as:

0.195 = X + 0.453 - 3X

0.195 = 0.453 - 2X

2X = 0.453 - 0.195

2X = 0.258

$X=\frac{0.258}{2}$

X = 0.129

So, there are 0.129 moles of methane in the mixture.

moles of propane = 0.151 - 0.129 = 0.022

mole fraction of methane = $\frac{moles of Methane}{total moles}$

mole fraction of methane = $\frac{0.129}{0.151}$

mole fraction of methane = 0.854

Hence, the mole fraction of methane gas in the original gas mixture is 0.854.

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2 years ago