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3 years ago

How many moles of Oxygen atoms are in a sample of 3.98 x 10^23 atoms?

2 answers:

8 0

$1mol\ \ \ \ \rightarrow \ \ \ 6,02*10^{23}\\
x \ \ \ \ \ \ \ \ \ \rightarrow \ \ \ 3,98*10^{23}\\\\
x=\frac{3,98*10^{23}*1mol}{6,02*10^{23}}\approx0,66mol$

san4es73 [151]3 years ago

7 0

Explanation:

As per the mole concept, there are $6.022 \times 10^{23} atoms$ present in 1 mole of a substance.

Mathematically, 1 mole = $6.022 \times 10^{23}$ atoms or molecules

Therefore, number of moles of oxygen atoms present in $3.98 \times 10^{23}$ atoms are calculated as follows.

$\frac{3.98 \times 10^{23} atoms}{6.022 \times 10^{23}atoms/mol}$

= 0.66 mol

Thus, we can conclude that there are 0.66 moles of oxygen atoms are in a sample of $3.98 \times 10^{23} atoms$ .

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How many atoms are in each of the following molecules:A. CO2B. N2C. CHCOOH

zlopas [31]

<h2>Answer:</h2>

A) 3 atoms - 1 atom of Carbon and 2 atoms of oxygen.

B) 2 atoms of Nitrogen.

C) 6 atoms - 2 Carbon atoms, 2 Hydrogen atoms, and 2 Oxygen atoms.

<h2>Explanations:</h2>

A molecule is a group of atoms bonded together, representing the smallest fundamental unit of a chemical compound. Molecules are made up of atoms.

According to the following information, we are to find the number of atoms in the given molecules.

A) For carbon dioxide CO₂, this molecule is made of 3 atoms - 1 atom of Carbon and 2 atoms of oxygen.

B) For the compound N₂, this molecule is made up of 2 atoms of Nitrogen.

C) For the compound CHCOOH, this molecule consists of 6 atoms - 2 Carbon atoms, 2 Hydrogen atoms, and 2 Oxygen atoms.

4 0

1 year ago

Fritz-Haber process

maks197457 [2]

Answer:

5×10⁵ L of ammonia (NH3)

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

N2 + 3H2 —> 2NH3

From the balanced equation above, we can say that:

3 L of H2 reacted to produce 2 L of NH3.

Finally, we shall determine the volume of ammonia (NH3) produced by the reaction of 7.5×10⁵ L of H2. This can be obtained as illustrated below:

From the balanced equation above,

3 L of H2 reacted to produce 2 L of NH3.

Therefore, 7.5×10⁵ L of H2 will react to produce = (7.5×10⁵ × 2)/3 = 5×10⁵ L of NH3.

Thus, 5×10⁵ L of ammonia (NH3) is produced from the reaction.

6 0

3 years ago

The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

MAVERICK [17]

Answer:

Approximately $2.46\; \rm mol$ .

Explanation:

Make use of the molar mass data ( $M({\rm C_2H_2}) = 26.04\; \rm g \cdot mol^{-1}$ ) to calculate the number of moles of molecules in that $64.0\; \rm g$ of $\rm C_2H_2$ :

$\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}$ .

Make sure that the equation for this reaction is balanced.

Coefficient of $\rm C_2H_2$ in this equation: $2$ .

Coefficient of $\rm H_2O$ in this equation: $2$ .

In other words, for every two moles of $\rm C_2H_2$ that this reaction consumes, two moles of $\rm H_2O$ would be produced.

Equivalently, for every mole of $\rm C_2H_2$ that this reaction consumes, one mole of $\rm H_2O$ would be produced.

Hence the ratio: $\displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1$ .

Apply this ratio to find the number of moles of $\rm H_2O$ that this reaction would have produced:

$\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}$ .

3 0

3 years ago

Determine the number of grams in each of the quantities<br><br> 1.39.0 x 1024 molecules Cl2

STatiana [176]

Mass of Cl₂ : 164.01 g

<h3>Further explanation</h3>

A mole is a number of particles(atoms, molecules, ions) in a substance

This refers to the atomic total of the 12 gr C-12 which is equal to 6.02.10²³, so 1 mole = 6.02.10²³ particles

Can be formulated :

N = n x No

N = number of particles

n = mol

No = 6.02.10²³ = Avogadro's number

mol Cl₂ :

$\tt n=\dfrac{N}{No}\\\\n=\dfrac{1.39.10^{24}}{6.02.10^{23}}\\\\n=2.31$

mass Cl₂(MW=71 g/mol) :

$\tt mass=mol\times MW\\\\mass=2.31\times 71=164.01$

8 0

3 years ago

For the reaction PC15 (8) PC13 (g) + Cl2 (g) K = 0.0454 at 261 °C. If a vessel is filled with these gases such that the initial

Fiesta28 [93]

Answer:

The correct answer is B.

The $K_{eq}$ is samller than $Q$ of the reaction . So,the reaction will shift towards the left i.e. towards the reactant side.

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given chemical reaction:

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$