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2 years ago

What is the percentage sodium in sodium oxide

Chemistry

2 answers:

Sav [38]2 years ago

5 0

Sodium oxide =-N220 i think

Alekssandra [29.7K]2 years ago

4 0

Sodium Oxide= Na2O
The formula mass of Na2O is (2x23) + 16 = 62g/mol

% Na= (46/62) x 100 = 74%
% O= (16/62) x 100 = 26%

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Which of the following would be most useful in trying to obtain procedural information to replicate an experiment previously pub

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Answer: Peer-reviewed journal article is the most useful because the information in them had been carefully scrutinized and aproved by people who are experts in that particular field.

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3 years ago

How many grams of copper (II) hydroxide can be prepared from 2.4 grams of copper (II) nitrate (Cu(NO3)2 ) and excess sodium hydr

andreyandreev [35.5K]

Answer:

percentage yield = 67%

Explanation:

Mass of Cu(NO₃)₂ = 15.25 g

Mass of NaOH = 12.75 g

Percentage yield = ?

Solution:

Cu(NO₃)₂ + 2NaOH → Cu(OH)₂ + 2NaNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

Number of moles = 15.25 g /187.56 g/mol

Number of moles = 0.08 mol

Moles of NaOH :

Number of moles = mass/ molar mass

Number of moles = 12.75 g / 40 g/mol

Number of moles = 0.32 mol

Now we will compare the moles of Cu(OH)₂ with NaOH and Cu(NO₃)₂. NaOH : Cu(OH)₂

2 : 1

0.32 : 1/2×0.32 = 0.16 mol

Cu(NO₃)₂ : Cu(OH)₂

1 : 1

0.08 : 0.08

The number of moles produced by Cu(NO₃)₂ are less so it will limiting reactant.

Mass of Cu(OH)₂:

Mass = number of moles × molar mass

Mass = 0.08 mol × 97.6 g/mol

Mass = 7.808 g

Theoretical yield = 7.808 g

Percent yield:

percentage yield = Actual yield/ theoretical yield × 100

percentage yield = 5.23 g/ 7.808 g × 100

percentage yield = 0.67 × 100

percentage yield = 67%

5 0

2 years ago

A 25.0-mL sample containing Cu2+ gave an instrument signal of 25.2 units (corrected for a blank). When exactly 0.500 mL of 0.027

irga5000 [103]

Answer:

The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.

Explanation:

The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:

$C = \frac{n}{V} \\0.0275 = \frac{n}{0.0005} \\$

n = 1.375x10⁻⁵ mol

The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):

1.375x10⁻⁵ mol _________ 19.9 units

x _________ 25.2 units

x = 1.741x10⁻⁵mol

Finally, we can calculate the Cu²⁺ concentration :

C = 1.741x10⁻⁵mol / 0.025 L

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7 0

3 years ago

At a certain temperature the vapor pressure of pure thiophene is measured to be . Suppose a solution is prepared by mixing of th

Lesechka [4]

Answer:

0.35 atm

Explanation:

It seems the question is incomplete. But an internet search shows me these values for the question:

" At a certain temperature the vapor pressure of pure thiophene (C₄H₄S) is measured to be 0.60 atm. Suppose a solution is prepared by mixing 137. g of thiophene and 111. g of heptane (C₇H₁₆). Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal."

Keep in mind that if the values in your question are different, your answer will be different too. However the methodology will remain the same.

First we calculate the moles of thiophene and heptane, using their molar mass:

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111 g heptane ÷ 100 g/mol = 1.11 moles heptane

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The mole fraction of thiophene is:

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2 years ago

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2 years ago