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3 years ago

Hiw many moles of 12c in 3.50g

Chemistry

1 answer:

Olenka [21]3 years ago

6 0

3.50 grams carbon 12 (1 mole C(12)/12.0 grams)(6.022 X 10^23/1 mole C(12))

= 1.76 X 10^23 atoms

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what would the molarity of a solution be if you took 10 mL of a 13M stock solution and made a 300 mL solution?

nignag [31]

Answer:

0.43M

Explanation:

8 0

3 years ago

A sample of propane(c3h8)has 3.84x10^24 H atoms.

deff fn [24]

Answer:

A) 14. 25 × 10²³ Carbon atoms

B) 34.72 grams

Explanation:

1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.

The sample has 3.84 × 10²⁴ H atoms.

If 8 atoms of Hydrogrn are present in 1 molecule of propane.

3.84 × 10²⁴ H atoms are present in

$\mathfrak{ \frac{3.8 }{8} \times 10 ^{24}}$

= 4.75 × 10²³ molecules of Propane.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

No. of Carbon atoms in 1 molecule of propane = 3

=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³

= 14.25 × 10²³ 

________________________________________

Gram Molecular Mass of Propane(C3H8)

= 3 × 12 + 8 × 1

= 36 + 8

= 44 g

1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.

=> 6.02 × 10²³ molecules of Propane weigh = 44 g

=> 4. 75 × 10²³ molecules of Propane weigh =

$\mathsf{ \frac{44 }{6.02 \times {10}^{23} } \times 4.75 \times {10}^{23} }$

$\mathsf{ = \frac{44 }{6.02 \times \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}} }$

$\mathsf{ = \frac{44 }{6.02 } \times 4.75 }$

= 34.72 g

8 0

3 years ago

For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80

Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

$A\rightleftharpoons B$

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

$K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%} =0.25$

Thus, by recalling the Van't Hoff's equation, we can write:

$ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )$

Hence, we solve for the enthalpy change as follows:

$\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }$

Finally, we plug in the numbers to obtain:

$\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}$

Learn more:

5 0

3 years ago

What do you mean by electronegativity ?

Kamila [148]

Answer:

Electronegativity is a measure of an atom's ability to attract shared electrons to itself. On the periodic table, electronegativity generally increases as you move from left to right across a period and decreases as you move down a group.

Explanation:

Hope it is helpful....

6 0

3 years ago

Read 2 more answers

Name the following compound:

Marrrta [24]

Answer:

3-methylhexane

Explanation:

The compound is 6 carbon atoms long, a hexane. It branches with a methyl group on the third carbon group.

This makes it a 3-methylhexane