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Anton [14]
3 years ago
11

How many moles are in 2.31*10^21 atoms of lead

Chemistry
1 answer:
Anon25 [30]3 years ago
6 0
Hey there!:

Molar mass Lead  ( Pb ) = 207.2 g/mol

Therefore:

1 mole Pb --------------------- 6.02*10²³ atoms
? moles Pb -------------------- 2.31*10²¹ atoms

moles Pb = ( 2.31*10²¹ ) * 1 / ( 6.02*10²³ ) = 

moles Pb = ( 2.31*10²¹ ) / ( 6.02*10²³ ) = 

=> 0.00383 moles of Pb

Hope this helps ! 
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What mass of KNO3 would be needed to produce 18.4 liters of oxygen gas, measured at 1.50 x 10^3 kPa and 15 degrees Celsius?
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Answer:-

2328.454 grams

Explanation:-

Volume V = 18.4 litres

Temperature T = 15 C + 273 = 288 K

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We know universal Gas constant R = 8.314 L KPa K-1 mol-1

Using the relation PV = nRT

Number of moles of oxygen gas n = PV / RT

Plugging in the values

n = (1.5 x 10^3 KPa ) x ( 18.4 litres ) / ( 8.314 L KPa K-1 mol-1 x 288 K)

n = 11.527 mol

Now the balanced chemical equation for this reaction is

2KNO3 --> 2KNO2 + O2

From the equation we can see that

1 mol of O2 is produced from 2 mol of KNO3.

∴ 11.527 mol of O2 is produced from 2 x 11.527 mol of KNO3.

= 23.054 mol of KNO3

Molar mass of KNO3 = 39 x 1 + 14 x 1 + 16 x 3 = 101 grams / mol

Mass of KNO3 = 23.054 mol x 101 gram / mol

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7 0
3 years ago
Choose the correct statement regarding the relative size of atoms:
Sonbull [250]

Answer:

The atoms on left side are larger than the atoms on the right side of the periodic table because those on the right have more proton's.

Explanation:

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But the electron's add to the same outer shell throughout the period , which means the effective nuclear charge increases which pulls the outer electrons toward's the nucleus and the size decreases.

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What is the difference between Hydrolysis and Hydration?​
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Mrrafil [7]

Answer:

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Explanation:

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2 years ago
The reducing agent in this catalytic hydrogenation reaction was molecular hydrogen (H2), which was produced in situ (in the reac
creativ13 [48]

Answer:

a)  After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) 4NaBH_{4} + 2HCl + 7H_{2}O ⇒ 16H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7} c) No H_{2} was not the limiting reactant.

Explanation:

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