Calculate the number of particles in 5.0 grams of NaCl.

Answers

Answer 1
Answer:

Answer:

5.2 × 10²² particles NaCl

General Formulas and Concepts:

Chemistry - Atomic Structure

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

5.0 g NaCl

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of Na - 22.99 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

Step 3: Convert

5.0 \ g \ NaCl((1 \ mol \ NaCl)/(58.44 \ g \ NaCl) )((6.022 \cdot 10^(23) \ particles \ NaCl)/(1 \ mol \ NaCl) ) = 5.15229 × 10²² particles NaCl

Step 4: Check

We are given 2 sig figs. Follow sig fig rules and round.

5.15229 × 10²² particles NaCl ≈ 5.2 × 10²² particles NaCl


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Answers

Answer:

left to right across a period when it decreases and when it increases top to bottom in a group,

hope i helped

Consider the balanced equation for the following reaction:7O2(g) + 2C2H6(g) → 4CO2(g) + 6H2O(l)
Determine the amount of CO2(g) formed in the reaction if 8.00 grams of O2(g) reacts with an excess of C2H6(g) and the percent yield of CO2(g) is 90.0%.

Answers

Answer: The amount of carbon dioxide formed in the reaction is 5.663 grams

Explanation:

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of oxygen gas = 8 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=(8g)/(32g/mol)=0.25mol

For the given chemical equation:

7O_2(g)+2C_2H_6(g)\rightarrow 4CO_2(g)+6H_2O(l)

By Stoichiometry of the reaction:

7 moles of oxygen gas produces 4 moles of carbon dioxide

So, 0.25 moles of oxygen gas will produce = (4)/(7)* 0.25=0.143mol of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.143 moles

Putting values in equation 1, we get:

0.143mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\n\n\text{Mass of carbon dioxide}=(0.143mol* 44g/mol)=6.292g

To calculate the experimental yield of carbon dioxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.292 g

Putting values in above equation, we get:

90=\frac{\text{Experimental yield of carbon dioxide}}{6.292g}* 100\n\n\text{Experimental yield of carbon dioxide}=(90* 6.292)/(100)=5.663g

Hence, the amount of carbon dioxide formed in the reaction is 5.663 grams

What is the most likely position for the hurricane indicated by the wind readings from the three weather stations shown?

Answers

Here's the answer, I remember doing this problem last year.

23.5 degrees north, 77 degrees west

How many sodium ions are in 1.4 kg of sodium chloride, NaCl?

Answers

Answer:

1.44 x 10²⁵ ions of Na⁺

Explanation:

Given parameters:

Mass of NaCl  = 1.4kg  = 1400g

Unknown:

Number of ions of sodium  = ?

Solution:

The compound NaCl in ionic form can be written as;

      NaCl →  Na⁺ + Cl⁻

In 1 mole of NaCl we have 1 mole of sodium ions

 Now, let us find the number of moles in NaCl;

  Number of moles  = (mass)/(molar mass)  

    Molar mass of NaCl  = 23 + 35.5 = 58.5g/mol

Number of moles  =  (1400)/(58.5)     = 23.93mol

 So;

   Since 1 mole of NaCl gives 1 mole of Na⁺  

    In 23.93 mole of NaCl will give 23.93 mole of Na⁺

1 mole of a substance  = 6.02 x 10²³ ions of a substance

  23.93 mole of a substance  =  6.02 x 10²³ x  23.93

                                                   = 1.44 x 10²⁵ ions of Na⁺

Round the following number to three significant figures: 156.5021 L

Answers

you round the 6 to a seven because there is a five following the decimal157

Consider the titration of a 73.9 mL sample of 0.13 M HC2H3O2 with 6.978 M NaOH. Ka(HC2H3O2) = 1.8x10-5 Determine the initial pH before any NaOH is added. Express your answer using two decimal places.Consider the titration of a 46.6 mL sample of 0.078 M HC2H3O2 with 1.135 M NaOH. Ka(HC2H3O2) = 1.8x10-5 Determine the volume of added base required to reach the equivalence point. Answer in units of milliliters.

Consider the titration of a 17.2 mL sample of 0.128 M HC2H3O2 with 0.155 M NaOH. Ka(HC2H3O2) = 1.8x10-5 Determine the pH at 0.46 mL of added base.

Answers

Answer:

1. pH = 2,82

2. 3,20mL of 1,135M NaOH

3. pH = 3,25

Explanation:

The buffer of acetic acid (HC₂H₃O₂) is:

HC₂H₃O₂ ⇄ H⁺ + C₂H₃O₂⁻

The reaction of HC₂H₃O₂ with NaOH produce:

HC₂H₃O₂ + NaOH → C₂H₃O₂⁻ + Na⁺ + H₂O

And ka is defined as:

ka = [H⁺] [C₂H₃O₂⁻] / [HC₂H₃O₂] = 1,8x10⁻⁵ (1)

1. When in the solution you have just 0,13M HC₂H₃O₂ the concentrations in equilibrium will be:

[H⁺] = x

[C₂H₃O₂⁻] = x

[HC₂H₃O₂] = 0,13 - x

Replacing in (1)

[x] [x] / [0,13-x] = 1,8x10⁻⁵

x² = 2,34x10⁻⁶ - 1,8x10⁻⁵x

x² - 2,34x10⁻⁶ + 1,8x10⁻⁵x  = 0

Solving for x:

x = - 0,0015 (Wrong answer, there is no negative concentrations)

x = 0,0015

As [H⁺] = x = 0,0015 and pH is -log [H⁺], pH of the solution is 2,82

2. The equivalence point is reached when moles of HC₂H₃O₂ are equal to moles of NaOH. Moles of HC₂H₃O₂ are:

0,0466L × (0,078mol / L) = 3,63x10⁻³ moles of HC₂H₃O₂

In a 1,135M NaOH, these moles are reached with the addition of:

3,63x10⁻³ moles × (L / 1,135mol) = 3,20x10⁻³L = 3,20mL of 1,135M NaOH

3. The initial moles of HC₂H₃O₂ are:

0,0172L × (0,128mol / L) = 2,20x10⁻³ moles of HC₂H₃O₂

As the addition of NaOH spent HC₂H₃O₂ producing C₂H₃O₂⁻. Moles of C₂H₃O₂⁻ are equal to moles of NaOH and moles of HC₂H₃O₂ are initial moles - moles of NaOH. That means:

0,46x10⁻³L NaOH × (0,155mol / L) = 7,13x10⁻⁵ moles of NaOH ≡ moles of C₂H₃O₂⁻

Final moles of HC₂H₃O₂ are:

2,20x10⁻³ - 7,13x10⁻⁵ = 2,2187x10⁻³ moles of HC₂H₃O₂

Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [C₂H₃O₂⁻] / [HC₂H₃O₂]

Where pka is -log ka = 4,74. Replacing:

pH = 4,74 + log₁₀ [7,13x10⁻⁵] / [2,2187x10⁻³ ]

pH = 3,25

I hope it helps!