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2 years ago

Which feature of the evaporating dish makes it effective in the evaporation process

Chemistry

2 answers:

Umnica [9.8K]2 years ago

5 0

Answer:

The precise fit on the ring stand. The ability to hold more substance than a test tube. A shallow bottom to increase the surface area of a liquid, leaving the solid behind.

Explanation:

Q and A!

How does an evaporating dish work?

Evaporating dishes are used to evaporate excess solvents – most commonly water – to produce a concentrated solution or a solid precipitate of the dissolved substance. ... The dish is heated with a Bunsen burner, until only stable precipitate remains, which contains the silica content.

What does a evaporating dish looks like?

The evaporating dish is a small bowl with a spout, usually made of porcelain or borosilicate glass. As its name suggests, it is commonly used to evaporate solvents in a sample. The evaporating dishes we encounter in a chemistry lab usually accommodate small samples.

<h2><u><em>Good Look On Your Assignment- Joshua Amachee :) </em></u></h2>

tigry1 [53]2 years ago

4 0

The precise fit on the ring stand. The ability to hold more substance than a test tube. A shallow bottom to increase the surface area of a liquid, leaving the solid behind.

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A tank holds 6.90 m3 of water. What is the volume in ft3? (3.28 ft = 1 m, Remember significant figures, but this answer does not

AleksAgata [21]

Volume of tank = $6.9 m^{3}$ (given)

Since, $1 m = 3.28 ft$

So,

$1 m^{3} = 35.287 ft^{3}$

For $6.9 m^{3}$ :

$6.9\times 35.287 ft^{3} = 243.4803 ft^{3}$

The significant rule for multiplication, states that the number of significant figures in the answer obtained by multiplication is determined by the value with the lowest number of significant digits.

Since, the minimum number of decimal places in the above multiplication operation is 1 so, the final result must be upto 1 decimal place only.

$243.4803 ft^{3} \simeq 243.5 ft^{3}$

Hence, volume in $ft^{3}$ is 243.5.

5 1

2 years ago

Read 2 more answers

For each of the following compounds, state whether it is ionic or covalent, and if it is ionic, write the symbols for the ions i

kupik [55]

Answer :

Covalent compound : It is defined as the compound which is formed by the sharing of electrons between the atoms forming a compound.

The covalent compound are usually formed when two non-metals react.

Ionic compound : It is defined as the compound which is formed when electron gets transferred from one atom to another atom.

All the polyatomic ions always form an ionic compound.

Polyatomic ions : It is a charged species that composed of two or more atoms and these charged species are bonded by the covalent bond.

For the given options:

Option A: $KClO_4$

This compound is formed by the combination of potassium, $K^+$ which is a metal and $ClO_4^-$ ion which is a polyatomic ion. Thus, it will form an ionic compound.

Option B: $Mg(C_2H_3O_2)_2$

This compound is formed by the combination of magnesium, $Mg^{2+}$ which is a metal and $C_2H_3O_2^{-}$ ion which is a polyatomic ion. Thus, it will form an ionic compound.

Option C: $H_2S$

Hydrogen and sulfur, both are non-metals and they will form a covalent compound.

Option D: $Ag_2S$

This compound is formed by the combination of silver, $Ag^{+}$ which is a metal and sulfur, $S^{2-}$ which is a non-metal. Thus, it will form an ionic compound.

Option E: $N_2Cl_4$

Nitrogen and chlorine, both are non-metals and they will form a covalent compound.

Option F: $Co(NO_3)_2$

This compound is formed by the combination of cobalt, $Co^{2+}$ which is a metal and $NO_3^{-}$ ion which is a polyatomic ion. Thus, it will form an ionic compound.

6 0

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 = $8.60g(\frac{1mol}{44g})$