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3 years ago

Tion 1 of 20

Chemistry

2 answers:

GarryVolchara [31]3 years ago

6 0

Answer:

None of these

Explanation:

The questions asks us to find the amount of sodium chloride formed (in grams) when 1.125 X 10^45 molecules of chlorine gas reacts with sodium.

We start by converting the molecules of chlorine gas to moles using Avogadro's number:

1.125 X 10^45 molecules Cl2 X 1 mol Cl2 / 6.022 X 10^23 = 1.87 X 10^21 mol

From here we can set up a BCA table, assuming that we have excess Sodium to react with: We have to assume that Cl2 is the limiting reactant.

2Na + Cl2 ----> 2NaCl

B . excess 1.87 X 10^21 0

C -3.74 X 10^21 -1.87 X 10^21 + 3.74 X 10^21

A X Amount 0 3.74 X 10^21

Now we have 3.74 X 10^21 moles of NaCl that we have to convert to grams:

We can use the molar mass of sodium chloride to accomplish this:

3.74 X 10^21 moles NaCl X 58.44 g NaCl / 1 mol NaCl = 2.18 X 10^23 g NaCl

Answer: None of these

finlep [7]3 years ago

5 0

Answer:

The mass of NaCl produced is 1.2 * 10^23 grams

Explanation:

Step 1: Data given

Number of molecules of Cl2 - gas = 1.25 * 10^45 molecules

Number of Avogadro = 6.022 * 10^23

Step 2: The balanced equation

2 Na + Cl2 → 2 NaCl

Step 3: Calculate moles of Cl2

Moles Cl2 = 1.25 *10^45 / 6.022*10^23

Moles Cl2 = 2.1 *10^21 moles

Step 4: Calculate moles NaCl

For 2.1 *10^21 moles Na we'll have 2.1*10^21 moles NaCl

Step 5: Calculate mass NaCl

Mass NaCl = moles NaCl * molar mass NaCl

Mass NaCl = 2.1*10^21 moles * 58.44 g/mol

Mass NaCl = 1.2 *10^23 grams

The mass of NaCl produced is 1.2 * 10^23 grams

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coldgirl [10]

Answer:

<h2>Density = 0.00026 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

<h3> $Density(\rho) = \frac{mass}{volume}$ </h3>

From the question

mass of air = 1.2 g

volume = 4,555 mL

Substitute the values into the above formula and solve for the density

That's

<h3> $Density = \frac{1.2}{4555}$ </h3>

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We have the final answer as

<h3>Density = 0.00026 g/mL</h3>

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2 years ago

What subatomic particle of the atom has no electric charge

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3 years ago

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lozanna [386]

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3 years ago

Calculate the molarity of sodium ion in a solution made

Arada [10]

Answer:

0.1035 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Sodium chloride will furnish Sodium ions as:

$NaCl\rightarrow Na^{+}+Cl^-$

Given :

For Sodium chloride :

Molarity = 0.288 M

Volume = 3.58 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 3.58×10⁻³ L

Thus, moles of Sodium furnished by Sodium chloride is same the moles of Sodium chloride as shown below:

$Moles =0.288 \times {3.58\times 10^{-3}}\ moles$

Moles of sodium ions by sodium chloride = 0.00103104 moles

Sodium sulfate will furnish Sodium ions as:

$Na_2SO_4\rightarrow 2Na^{+}+SO_4^{2-}$

Given :

For Sodium sulfate :

Molarity = 0.001 M

Volume = 6.51 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 6.51 ×10⁻³ L

Thus, moles of Sodium furnished by Sodium sulfate is twice the moles of Sodium sulfate as shown below:

$Moles =2\times 0.001 \times {6.51\times 10^{-3}}\ moles$

Moles of sodium ions by Sodium sulfate = 0.00001302 moles

Total moles = 0.00103104 moles + 0.00001302 moles = 0.00104406 moles

Total volume = 3.58 ×10⁻³ L + 6.51 ×10⁻³ L = 10.09 ×10⁻³ L

Concentration of sodium ions is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{Na^+}=\frac{0.00104406}{10.09\times 10^{-3}}$

<u> The final concentration of sodium anion = 0.1035 M</u>

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3 years ago