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REY [17]
3 years ago
8

How many molecules are present in a drop of ethanol, c2h5oh, of mass 2.3*10^-2.3g? (NA=6.0*10^23mol^-1)

Chemistry
1 answer:
sdas [7]3 years ago
5 0

Answer:

3.0 × 10²⁰ molecules

Explanation:

Given data:

Mass of ethanol = 2.3 × 10⁻²°³ g

Number of molecules = ?

Solution:

Number of moles of ethanol:

Number of moles = mass/ molar mass

Number of moles = 2.3 × 10⁻²°³ g / 46.07 g/mol

Number of moles = 0.05 × 10⁻²°³ mol

Number of molecules:

One mole = 6.022 × 10²³ molecules

0.05 × 10⁻²°³ mol  ×  6.022 × 10²³ molecules / 1 mol

0.30 × 10²⁰°⁷ molecules

3.0 × 10¹⁹°⁷ molecules which is almost equal to 3.0 × 10²⁰ molecules.

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A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) →
vova2212 [387]

Answer:

Copper (II) sulfate

Explanation:

Given reaction is

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)

Amount of aluminum = 1·25 g

Amount of copper (II) sulfate = 3·28 g

Atomic weight of Al = 26 g

Molecular weight of CuSO4 ≈ 159·5

Number of moles of Al = 1·25 ÷ 26 = 0·048

Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021

From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required

So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required

For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required

∴ Number of moles of copper (ll) sulfate required = 0·072

But we have only 0·021 moles of copper (ll) sulfate

As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate

∴ The limiting reactant is copper (ll) sulfate

7 0
3 years ago
How many milliliters of a stock solution of 6.20 M HNO3 would you have to use to prepare 0.130 L of 0.470 M HNO3?
Mandarinka [93]

Answer:

9.85

Explanation:

M1V1 =M2V2

6.20×v1= 0.470×0.130

v1 = ( 0.470 × 0.130 ) ÷ 6.20

v1 = 0.0098 L × 1000

V1 = 9.8 ml

6 0
2 years ago
C-12 has 6 protons.<br> How many neutrons does C-13 have?<br> A. 6<br> B. 7<br> C. 13<br> D. 19
atroni [7]

Answer:

B-7

Explanation:

13-6=7

8 0
3 years ago
Read 2 more answers
3. Matter with a composition that is always the same is a(n)<br> substance<br> mixture<br> compound
Andre45 [30]

Answer:

substance

Explanation:

A pure substance is a form of matter that has a constant composition (meaning it's the same everywhere) and properties that are constant throughout the sample (meaning there is only one set of properties such as melting point, color, boiling point, etc

7 0
3 years ago
What is the molarity of a solution that contains 122g of MgSO4 n 3.5L of solution?​
Semmy [17]

Answer:

0.29mol/L or 0.29moldm⁻³

Explanation:

Given parameters:

Mass of MgSO₄ = 122g

Volume of solution = 3.5L

Molarity is simply the concentration of substances in a solution.

Molarity = number of moles/ Volume

>>>>To calculate the Molarity of MgSO₄ we find the number of moles using the mass of MgSO₄ given.

Number of moles = mass/ molar mass

Molar mass of MgSO₄:

Atomic masses: Mg = 24g

S = 32g

O = 16g

Molar mass of MgSO₄ = [24 + 32 + (16x4)]g/mol

= (24 + 32 + 64)g/mol

= 120g/mol

Number of moles = 122/120 = 1.02mol

>>>> From the given number of moles we can evaluate the Molarity using this equation:

Molarity = number of moles/ Volume

Molarity of MgSO₄ = 1.02mol/3.5L

= 0.29mol/L

IL = 1dm³

The Molarity of MgSO₄ = 0.29moldm⁻³

8 0
3 years ago
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