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3 years ago

12

Spell out the full name of the compound.

2 answers:

djverab [1.8K]3 years ago

7 0

C3H8 Propane ! Would be ur answer

andre [41]3 years ago

5 0

Answer:

it is propane

C3H8 it is propane

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What is the only intermolecular force present in nonpolar compounds?

Gnom [1K]

The intermolecular force of attraction between non-polar molecules is called as dispersion force . Another trem for dispersion force is London dispersion force.

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4 years ago

How will you test for the gas which is liberated when hcl reacts with an active metal ?

DENIUS [597]

When HCl reacts with a metal, hydrogen gas will be evolved. To test this gas, insert a burning splinter into the outlet of gas, the flame will be extinguished with a pop sound. This will confirm the gas is hydrogen.

8 0

3 years ago

We have 548 grams of CaCl2. How many molecules of CaCl2 do we have? (The molar mass of CaCl2 is 110.98 g/mol)

nekit [7.7K]

Answer:

Try everything

2.97 x 10^24 molecules

3.66 x 10^28 molecules

9.1 x 10^22 molecules

4.37 x 10^22 molecules

Explanation:

3 0

3 years ago

The PH of a solution of Hcl is 2.find out the amount of acid present in a litre of the solution

Scrat [10]

Answer:

The solution is 10^-2 or 0.01M in HCl.

Explanation:

meaning of pH is "power of hydrogen".

what is the molar concentration of a HCl solution with pH=2?

Let say pH=2

[H+]=10^-2M

HCL is a strong acid that dissociates completely:

[H+]=[HCL]

Therefore solution is 10^-2 or 0.01M in HCL.

5 0

3 years ago

How many grams of sodium fluoride should be added to 300. mL of 0.0310 M of hydrofluoric acid to produce a buffer solution with

Kisachek [45]

Answer : The mass of sodium fluoride added should be 0.105 grams.

Explanation : Given,

The dissociation constant for HF = $K_a=6.8\times 10^{-4}$

Concentration of HF (weak acid)= 0.0310 M

First we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (6.8\times 10^{-4})$

$pK_a=4-\log (6.8)$

$pK_a=3.17$

Now we have to calculate the concentration of NaF.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[NaF]}{[HF]}$

Now put all the given values in this expression, we get:

$2.60=3.17+\log (\frac{[NaF]}{0.0310})$

$[NaF]=0.00834M$

Now we have to calculate the moles of NaF.

$\text{Moles of NaF}=\text{Concentration of NaF}\times \text{Volume of solution}=0.00834M\times 0.300L=0.0025mole$

Now we have to calculate the mass of NaF.

$\text{Mass of }NaF=\text{Moles of }NaF\times \text{Molar mass of }NaF=0.0025mole\times 42g/mole=0.105g$

Therefore, the mass of sodium fluoride added should be 0.105 grams.

4 0

4 years ago