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__

__ = 5.15229 × 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

The reform reaction between steam and gaseous methane () produces "synthesis gas," a mixture of carbon monoxide gas and dihydrogen gas. Synthesis gas is one of the most widely used industrial chemicals, and is the major industrial source of hydrogen. Suppose a chemical engineer studying a new catalyst for the reform reaction finds that liters per second of methane are consumed when the reaction is run at and . Calculate the rate at which dihydrogen is being produced. Give your answer in kilograms per second. Round your answer to significant digits..

It is the outermost layer of the solid portion of earth

Which one of the following combinations cannot function as a buffer solution?A) HCN and KCNB) NH3 and (NH4)2SO4C) HNO3 and NaNO3D) HF and NAFE) HNO2 and NaNO2

At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressure of He would give a solubility of 0.730 M

A sample of gas has an initial volume of 3.00 L and an initial pressure of 2.14 atm. If the volume expands to 8.15 L, what is the final pressure?a. 0.855 atm b. 0.788 atm c. 3.49 atm d. 5.81 atm

It is the outermost layer of the solid portion of earth

Which one of the following combinations cannot function as a buffer solution?A) HCN and KCNB) NH3 and (NH4)2SO4C) HNO3 and NaNO3D) HF and NAFE) HNO2 and NaNO2

At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressure of He would give a solubility of 0.730 M

A sample of gas has an initial volume of 3.00 L and an initial pressure of 2.14 atm. If the volume expands to 8.15 L, what is the final pressure?a. 0.855 atm b. 0.788 atm c. 3.49 atm d. 5.81 atm

largest atomic radii.

**Answer:**

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

hope i helped

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%.

**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:**

** .....(1)**

Given mass of oxygen gas = 8 g

Molar mass of oxygen gas = 32 g/mol

**Putting values in equation 1, we get:**

**For the given chemical equation:**

**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 = 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:**

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

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.292 g

**Putting values in above equation, we get:**

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

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

23.5 degrees north, 77 degrees west

**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 =

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

Number of moles = = 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⁺**

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

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.

**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!