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3 years ago

How many molecules are there in 45.0 grams of CH4?

Chemistry

1 answer:

user100 [1]3 years ago

5 0

Answer: $16.93\times 10^{23}\ \text{molecules}$

Explanation:

Given

Mass of $CH_4$ $m=45\ gm$

Molar mass of $CH_4$ is $12+4\times1=16$

No of moles of $CH_4$ $=\dfrac{\text{Mass}}{\text{Molar mass}}=\frac{45}{16}$

No of Molecules in 45gm of $CH_4$ is

$\Rightarrow N_A\times\frac{45}{16}=6.022\times 10^{23}\times \frac{45}{16}\\\Rightarrow 16.93\times10^{23}\ \text{molecules}$

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What is molarity of 47.0 g KCl dissolved in enough water to give 375 mL of solution?

natita [175]

This question provides us –

Weight of $\bf KCl$ is = 47 g
Volume, V = 375 mL

__________________________________________

Molar Mass of $\bf KCl$ –

$\qquad$ $\twoheadrightarrow\bf 39.0983 \times 35.453$

$\qquad$ $\twoheadrightarrow\bf 74.5513$

<u>Using formula</u> –

$\qquad$ $\purple{\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{ W\times 1000}{MV}}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{ 47 \times 1000}{74.5513\times 375}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{47000}{27956.7375}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = \cancel{\dfrac{47000}{27956.7375}}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = 1.68117M$

$\qquad$ $\pink{\twoheadrightarrow\bf Molarity _{(Solution)} = 1.7M}$

Henceforth, Molarity of the solution is = 1.7M

___________________________________________

6 0

2 years ago

Find the percentage composition of a compound that contains 1.94 g of carbon, 0.48 g of hydrogen, and 2.58 g of sulfur.

Svetradugi [14.3K]

Answer : The percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

Explanation :

To calculate the percentage composition of element in sample, we use the equation:

$\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100$

Given:

Mass of carbon = 1.94 g

Mass of hydrogen = 0.48 g

Mass of sulfur = 2.58 g

First we have to calculate the mass of sample.

Mass of sample = Mass of carbon + Mass of hydrogen + Mass of sulfur

Mass of sample = 1.94 + 0.48 + 2.58 = 5.0 g

Now we have to calculate the percentage composition of a compound.

$\%\text{ composition of carbon}=\frac{1.94g}{5.0g}\times 100=38.8\%$

$\%\text{ composition of hydrogen}=\frac{0.48g}{5.0g}\times 100=9.6\%$

$\%\text{ composition of sulfur}=\frac{2.58g}{5.0g}\times 100=51.6\%$

Hence, the percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

3 0

3 years ago

The element lanthanum has two stable isotopes, lanthanum-138 with an atomic mass of 137.9 AMU and lanthanum-139 with an atomic m

dmitriy555 [2]

Answer:

A. there is an isotope of lanthanum with an atomic mass of 138.9

Explanation:

By knowing the different atomic masses of both Lanthanum atoms, we can not tell anything about their occurence in nature. Therefore, all the last three options are incorrect. Because, the atomic mass does not tell anything about the availability or natural abundance of an element.

Now, the isotopes of an element are those elements, which have same number of electrons and protons as the original element, but different number of neutrons. Therefore, they have same atomic number but, different atomic weight or atomic masses.

Hence, by looking at an elements having same atomic number, but different atomic masses, we can identify them as isotopes.

Thus, the correct option is:

<u>A. there is an isotope of lanthanum with an atomic mass of 138.9.</u>

7 0

3 years ago

scandium47 has a half-life of 35s. suppose you have a 45g sample of scadium 47 how much of the sample remains unchanged after 14

Mars2501 [29]

The much of the sample that would remain unchanged after 140 seconds is 2.813 g

Explanation

Half life is time taken for the quantity to reduce to half its original value.

if the half life for Scandium is 35 sec, then the number of half life in 140 seconds

=140 sec/ 35 s = 4 half life

Therefore 45 g after first half life = 45 x1/2 =22.5 g

22.5 g after second half life = 22.5 x 1/2 =11.25 g

11.25 g after third half life = 11.25 x 1/2 = 5.625 g

5.625 after fourth half life = 5.625 x 1/2 = 2.813

therefore 2.813 g of Scandium 47 remains unchanged.

4 0

2 years ago

A solution is prepared by mixing 200.0 g of water, H2O, and 300.0 g of

VikaD [51]

Answer:

Mole Fraction (H₂O) = 0.6303

Mole Fraction (C₂H₅OH) = 0.3697

Explanation:

(Step 1)

Calculate the mole value of each substance using their molar masses.

Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol

Molar Mass (H₂O): 18.014 g/mol

200.0 g H₂O 1 mole
--------------------- x ------------------ = 11.10 moles H₂O
18.014 g

Molar Mass (C₂H₅OH): 2(12.011 g/mol) + 6(1.008 g/mol) + 15.998 g/mol

Molar Mass (C₂H₅OH): 46.068 g/mol

300.0 g C₂H₅OH 1 mole
---------------------------- x -------------------- = 6.512 moles C₂H₅OH
46.068 g

(Step 2)

Using the mole fraction ratio, calculate the mole fraction of each substance.

moles solute
Mole Fraction = ------------------------------------------------
moles solute + moles solvent

11.10 moles H₂O
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH

Mole Fraction (H₂O) = 0.6303

6.512 moles C₂H₅OH
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH

Mole Fraction (C₂H₅OH) = 0.3697

7 0

1 year ago