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miskamm [114]
4 years ago
11

How many atoms are in 10 moles of platinum

Chemistry
2 answers:
Karo-lina-s [1.5K]4 years ago
8 0
6.022x10^24 atoms


Where moles = # of atoms/ Avogadro's constant

10 = x/ 6.022x10^23

x = 6.022x10^24
Finger [1]4 years ago
7 0
10 mols platinum * 6.02*10^23 atoms = 6.02*10^23 atoms in platinum 
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Difference between cold process soap making and hot process soap making​
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7 0
3 years ago
Determine the maximum amount of Fe that was produced during the experiment. Explain how you determined this amount
VladimirAG [237]

The number of moles of iron produced can be obtained by the use of  stoichiometry.

<h3>What is stoichiometry?</h3>

The term stoichiometry has to do with mass - mole or mass - volume relationships. Stoichiometric relationships could be used to determine the amount of substance.

Now the reaction equation as well as other details are not shown in this question hence a numerical anwer can not be provided. As such, we can know that the balanced reaction equation can always be used to obtain the amount of iron by stoichiometry.

Learn more about stoichiometry: brainly.com/question/9743981

6 0
2 years ago
The pressure of a 5.0 L sample of gas changes from 3.0 atm to 10.0 atm while the temperature remains constant. What is the gas’s
lina2011 [118]

Answer:

<h2>The answer is 1.5 L</h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question we have

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8 0
3 years ago
Find the mass of 3.27 x 10^23 molecules of H2SO4. Use 3 significant digits<br> and put the units.
marta [7]

Answer:

Approximately 53.3\; \rm g.

Explanation:

Lookup Avogadro's Number: N_{\rm A} = 6.02\times 10^{23}\; \rm mol^{-1} (three significant figures.)

Lookup the relative atomic mass of \rm H, \rm S, and \rm O on a modern periodic table:

  • \rm H: 1.008.
  • \rm S: 32.06.
  • \rm O: 15.999.

(For example, the relative atomic mass of \rm H is 1.008 means that the mass of one mole of \rm H\! atoms would be approximately 1.008\! grams on average.)

The question counted the number of \rm H_2SO_4 molecules without using any unit. Avogadro's Number N_{\rm A} helps convert the unit of that count to moles.

Each mole of \rm H_2SO_4 molecules includes exactly (1\; {\rm mol} \times N_\text{A}) \approx 6.02\times 10^{23} of these \rm H_2SO_4 \! molecules.

3.27 \times 10^{23} \rm H_2SO_4 molecules would correspond to \displaystyle n = \frac{N}{N_{\rm A}} \approx \frac{3.27 \times 10^{23}}{6.02 \times 10^{23}\; \rm mol^{-1}} \approx 0.541389\; \rm mol of such molecules.

(Keep more significant figures than required during intermediary steps.)

The formula mass of \rm H_2SO_4 gives the mass of each mole of \rm H_2SO_4\! molecules. The value of the formula mass could be calculated using the relative atomic mass of each element:

\begin{aligned}& M({\rm H_2SO_4}) \\ &= (2 \times 1.008 + 32.06 + 4 \times 15.999)\; \rm g \cdot mol^{-1} \\ &= 98.702\; \rm g \cdot mol^{-1}\end{aligned}.

Calculate the mass of approximately 0.541389\; \rm mol of \rm H_2SO_4:

\begin{aligned}m &= n \cdot M \\ &\approx 0.541389\; \rm mol \times 98.702\; \rm g \cdot mol^{-1}\\ &\approx 53.3\; \rm g\end{aligned}.

(Rounded to three significant figures.)

6 0
3 years ago
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