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3 years ago

How many hydrogen atoms are contained in 0.40 mL of water, given that the density of water is 1 g/ml?

Chemistry

1 answer:

Paladinen [302]3 years ago

3 0

Answer:

2.676e22 atoms of Hydrogen

Explanation:

Knowing the density of water is 1g/ml, 1(g/ml)*0.4ml=0.4 grams of H2O. Knowing that there are 16 grams of H2O in 1 mole of H2O, we can set up a unit conversion of 0.4 Grams H2O * $\frac{1 mol H_2O}{18 grams H_2O}*\frac{2 mol H}{1 mol H_2O} *\frac{6.022e23 atoms of H}{1 mol H}$ =

2.676e22 atoms of Hydrogen.

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Suppose a solution is prepared by dissolving 15.0 g NaOH in 0.150 L of 0.250 M nitric acid. What is the final concentration of O

Kipish [7]

Answer:

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

Explanation:

Moles of NaOH = $\frac{15.0 g}{40 g/mol}=0.375 mol$

Molarity of the nitric acid solution = 0.250 M

Volume of the nitric solution = 0.150 L

Moles of nitric acid = n

$Molarity=\frac{Moles}{Volume(L)}$

$n=0.250 M\times 0.150 L=0.0375 mol$

$NaOH+HNO_3\rightarrow NaNO_3+H_2O$

According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :

$\frac{1}{1}\times 0.0375 mol$ of NaOH

Moles of NaOH left unreacted in the solution =

= 0.375 mol - 0.0375 mol = 0.3375 mol

$NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)$

1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.

Then 0.3375 moles of NaOH will give :

$1\times 0.3375 moles=0.3375 mol$ of hydroxide ion

The molarity of hydroxide ion in solution ;

$=\frac{0.3375 mol}{0.150 L}=2.25 M$

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

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3 years ago

A prototype is usually different from the final product

Assoli18 [71]

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3 years ago

2. What if hydrogen were left over?

valentina_108 [34]

Answer: limiting reactant (or limiting reagent): The reactant that determines the amount of product that can be formed in a chemical reaction.

Explanation: Your welcome buddy.

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2 years ago

Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 1.000 mg of caffeine

Yuliya22 [10]

Answer:

194 g/mol

Explanation:

1) Content of C:

All the C atoms in the 1.000 mg of caffeine will be found in the 1.813 mg of CO₂.

Mass of C in 1.813 mg of CO₂

Since the atomic mass of C is 12.01 g/mol and the molar of of CO₂ is 44.01, there are 12 mg of C in 44 mg of CO₂ and you can set the proporton:

12.01 mg C / 44.01 mg CO₂ = x / 1.813 g CO₂

⇒ x = 1.813 × 12.01 / 44.01 g of C = 0.49475 mg of C

Number of moles of C

number of moles = mass in g / atomic mass = 0.49475×10⁻³ g / 12.01 g/mol = 4.1195×10⁻⁵ moles = 0.041195 milimol

2) Content of H

All the H atoms in the 1.000 mg of caffeine will be found in the 0.4639 mg of H₂O

Mass of H in 0.4639 mg of H₂O

Since the atomic mass of H is 1.008 g/mol and the molar of of H₂O is 18.015 g/mol, there are 2×1.008 mg of H in 18.015 mg of H₂O and you can set the proporton:

2×1.008 mg H / 18.015 mg H₂O = x / 0.4639 mg H₂O

⇒ x = 0.4639 mg H₂O × 2 × 1.008 mg H / 18.015 mg H₂O = 0.051913 mg H

Number of moles of H

number of moles = mass in g / atomic mass = 0.051913 ×10⁻³ g / 1.008 g/mol = 5.1501× 10⁻⁵ moles = 0.051501 milimol

3) Content of N

All the N atoms in the 1.000 mg of caffeine will be found in the 0.2885 mg of N₂

Mass of N in 0.2885 mg of N₂ is 0.2885 mg

Number of moles of N

number of moles = mass in g / atomic mass = 0.2885 ×10⁻³ g / 14.007 g/mol = 2.0597× 10⁻⁵ moles = 0.020597 milimol

4) Content of O

The mass of O is calculated by difference:

Mass of O = mass of sample - mass of C - mass of H - mass of N

Mass of O = 1.000 mg - 0.49475 mg C - 0.051913 mg H - 0.2885 mg N

Mass of O = 0.1648 mg

Moles of O = 0.1648 × 10 ⁻³ g / 15.999 g/mol = 1.0303×10⁻⁵ mol = 0.01030 milimol

5) Ratios

Divide every number of mililmoles by the smallest number of milimoles:

C: 0.041195 / 0.01030 = 4
H: 0.051501 / 0.01030 = 5
N: 0.020597 / 0.01030 = 2

O: 0.01030 / 0.01030 = 1

C: 4
H: 5
N: 2
O: 1

6) Empirical formula:

C₄H₅N₂O₁

7) Calculate the approximate mass of the empirical formula:

4 × 12 + 5 × 1 + 2 × 14 + 1 × 16 = 97 g/mol

So, since that number is not between 150 and 200 g/mol, multiply by 2: 97 × 2 = 194, which is between 150 and 200.

Thus, the estimate is 194 g/mol

7 0

3 years ago