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3 years ago

6

You want to place a piece of glass tubing into a ruber stopper after the tubing has been fire polished and cooled. This is best

done by

1 answer:

igor_vitrenko [27]3 years ago

4 0

Answer:

All of the above

Explanation:

This is correct

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A compound's properties are different than the properties of the elements that make it up.

yan [13]

The answer is True because elements in a compound combine and become an entirely different substance with its own unique properties.

7 0

3 years ago

Use the following equation to answer the questions below:

Gala2k [10]

Explanation:

The equation of the reaction is given as;

Be + 2HCl → BeCl2 + H2

What is the mass of beryllium required to produce 25.0g of beryllium chloride?

1 mol of Be produces 1 mol of BeCl2

Converting to mass;

Mass = Molar mass * Number of moles

9.01g of Be produces 79.92g of BeCl2

xg of Be produces 25g of BeCl2

Solving for x;

x = 25 * 9.01 / 79.92

x = 2.82 g

What is the mass of hydrochloric acid required to produce 25.0g of beryllium chloride? g

Converting 25.0g of beryllium chloride to moles;

Number of moles = Mass / Molar mass

Number of moles = 25 / 79.92 = 0.3128 mol

2 mol of HCl produces 1 mol of BeCl2

x mol of HCl would produce 0.3128 mol of BeCl2

solving for x;

x = 0.3128 * 2 = 0.6256 mol

Converting to mass;

Mass = 0.6256 * 36.5 = 22.83 g

What is the mass of hydrogen gas produced when 25.0g of beryllium chloride is also produced? g

25g of BeCl2 = 0.3128 mol of BeCl2

From the equation;

1 mol of H2 is produced alongside 1 mol of BeCl2

This means;

0.3128 mol of H2 would also be produced alongside 0.3128 mol of BeCl2

Mass = Number of moles * Molar mass

Mass = 0.3128mol * 2.0159 g/mol = 0.6306 g

3 0

3 years ago

Chất nào sau đây là chất điện li mạnh trong H2O

erica [24]

Answer:

những lựa chọn ở đâu?

Explanation:

Hy vọng nó sẽ giúp ích;>

3 0

3 years ago

Read 2 more answers

In the context of small molecules with similar molar masses, arrange these intermolecular forces by strength (hydrogen bonding -

Katarina [22]

<h2>Answer:</h2>

Arrangement of inter molecular forces from strongest to weakest.

Hydrogen bonding
Dipole-dipole interactions
London dispersion forces.

<h3>Explanation:</h3>

Intermolecular forces are defined as the attractive forces between two molecules due to some polar sides of molecules. They can be between nonpolar molecules.

Hydrogen bonding is a type of dipole dipole interaction between the positive charge hydrogen ion and the slightly negative pole of a molecule. For example H---O bonding between water molecules.

Dipole dipole interactions are also attractive interactions between the slightly positive head of one molecule and the negative pole of other molecules.

But they are weaker than hydrogen bonding.

London dispersion forces are temporary interactions caused due to electronic dispersion in atoms of two molecules placed together. They are usually in nonpolar molecules like F2, I2. they are weakest interactions.

5 0

3 years ago

Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t

Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

$A=\log \frac{I_o}{I}$

$\log \frac{I_o}{I}=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution = $3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}$

l = path length = 2.5 mm = 0.25 cm

$I_o$ = incident light

$I$ = transmitted light

$\epsilon$ = molar absorptivity coefficient = $855dm^3mol^{-1}cm^{-1}$

Now put all the given values in the above formula, we get:

$\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)$

$\log \frac{I_o}{I}=0.6947$

$\frac{I_o}{I}=10^{0.6947}=4.951$

If we consider $I_o$ = 100

then, $I=\frac{100}{4.951}=20.198$

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light $I_A$ = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

$\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100$

$\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%$

Therefore, the percentage reduction in intensity is 79.80 %

3 0

4 years ago