2. Calculate the value (calories) for the heat of solution of 2.25 g of NaOH.

3. Calculate the number of calories that would be produced if one mole of sodium hydroxide was dissolved. (ΔHsolnNaOH)

Answer:

**Answer:**

**For 1:**** The molarity of sodium hydroxide solution is 0.375 M**

**For 2:**** The amount of heat absorbed by solution is 1674.75 Cal**

**For 3:**** The enthalpy change of the reaction when 1 mole of NaOH is dissolved is 1674.75 Cal**

**Explanation:**

**For 1:**

**To calculate mass of a substance, we use the equation:**

Density of water = 1 g/mL

Mass of water = 150.00 g

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

**To calculate the molarity of solution, we use the equation:**

**We are given:**

Mass of solute (sodium hydroxide) = 2.25 g

Molar mass of sodium hydroxide = 40 g/mol

Volume of solution = 150.00 mL

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

**Hence, the molarity of sodium hydroxide solution is 0.375 M**

**For 2:**

**To calculate the amount of heat absorbed, we use the equation:**

where,

m = mass of solution = (2.25 + 150) = 152.25 g

c = specific heat capacity of water = 1 Cal/g.°C

= change in temperature = 11°C

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

**Hence, the amount of heat absorbed by solution is 1674.75 Cal**

**For 3:**

**To calculate the enthalpy change of the reaction, we use the equation:**

where,

= amount of heat absorbed = 16

74.75 Cal

n = number of moles = 1 mole

= enthalpy change of the reaction

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

**Hence, the enthalpy change of the reaction when 1 mole of NaOH is dissolved is 1674.75 Cal**

Answer:
### Final answer:

### Explanation:

### Learn more about Thermodynamics and Stoichiometry here:

The molarity of the sodium hydroxide is 0.375 M. The heat of solution of the sodium hydroxide is -1650 cal, and the heat of solution per mole of sodium hydroxide is -29333.33 cal/mol.

To answer your questions, we first need to convert the mass of the **sodium hydroxide** (NaOH) to moles. Sodium hydroxide has a molar mass of approximately 40 g/mol, so 2.25 g is 0.05625 mol.

1. The **molarity** of the **solution **is the number of moles of solute per liter of solution. Given that the solution was made up in 150.00 g of **water**, which is approximately 0.150 L (since the density of water is approximately 1 g/mL), the molarity is 0.05625 mol / 0.150 L = 0.375 M.

2. The heat of solution can be calculated using the equation q = m * c * ΔT, where m is the mass of the water, c is the specific heat capacity of the water (approximately 1 cal/g°C), and ΔT is the change in temperature. Plugging in the known values, q = 150.00 g * 1 cal/g°C * 11°C = 1650 cal. This is the heat absorbed by the water and so the heat of solution of NaOH is -1650 cal (as the process of dissolving is exothermic).

3. The heat of solution per mole of sodium hydroxide can be calculated by dividing the total heat of solution by the number of moles of sodium hydroxide. So ΔHsoln NaOH = -1650 cal / 0.05625 mol = -29333.33 cal/mol.

#SPJ3

Describe how you would make 250 ml of a 3 M solution of sodium acetate (NaOAc = 82.03 g/mol). First figure out how much sodium acetate you would need, then describe how you would make the solution if you were given a bottle of solid sodium acetate, a volumetric flask, and DI water.

You have been injured in the laboratory (cut,burn,etc) first you should

Calculate the molarity of each solution.a. 0.38 mol of lino3 in 6.14 l of solutionb. 72.8 g c2h6o in 2.34 l of solutionc. 12.87 mg ki in 112.4 ml of solution

When NaOH is added to water, the (OH) = 0.04 M. What is the [H30*]?What is the PH of the solution?

Balance the following redox reaction occurring in an acidic solution. The coefficient of Mn2+(aq) is given. Enter the coefficients (integers) into the cells before each substance below the equation. ___ AsO2−(aq) + 3 Mn2+(aq) + ___ H2O(l) \rightarrow→ ___ As(s) + ___ MnO4−(aq) + ___ H+(aq)

You have been injured in the laboratory (cut,burn,etc) first you should

Calculate the molarity of each solution.a. 0.38 mol of lino3 in 6.14 l of solutionb. 72.8 g c2h6o in 2.34 l of solutionc. 12.87 mg ki in 112.4 ml of solution

When NaOH is added to water, the (OH) = 0.04 M. What is the [H30*]?What is the PH of the solution?

Balance the following redox reaction occurring in an acidic solution. The coefficient of Mn2+(aq) is given. Enter the coefficients (integers) into the cells before each substance below the equation. ___ AsO2−(aq) + 3 Mn2+(aq) + ___ H2O(l) \rightarrow→ ___ As(s) + ___ MnO4−(aq) + ___ H+(aq)

**Answer:**

q

**Explanation:**

**Explanation:**

Law of conservation of mass states that mass can neither be created nor it can be destroyed but it can be transformed into one form to another.

Similarly, law of conservation of energy states that energy can neither be created nor it can be destroyed as it can only be transformed from one form to another.

In modern view of matter and energy, is the law of mass conservation still relevant to chemical reactions as follows.

**For example, **

Atomic mass of Na = 23

Atomic mass of Cl = 35

Hence, mass of total number of reactants is calculated as follows.

[(2 \times 23) + (35 \times 2)] g/mol = 116 g/mol

Mass of total number of products is calculated as follows.

[2 \times (23 + 35)] = 116 g/mol

Thus, **it is proved that in our modern view of matter and energy, is the law of mass conservation still relevant to chemical reactions.**

**Answer:**

Which of the following is a possible set of quantum numbers for a 3p orbital?

Since you are dealing with a 3p-orbital, your principal and angular momentum quantum numbers will be n=3 and l=1 .

3.3 x 10^-25 atoms

1.01 x 10^-3 moles

n = 0.20 mol

N

N = n × 6.02 × 10²³ atoms/mol

N = 0.2 mol × 6.02 × 10²³ atoms/mol

__N = 1.20 × 10²³ atoms__

Therefore, there are **1.20 × 10²³** gold atoms in 0.2 mol of a gold sample.

#ILoveChemistry

#ILoveYouShaina

BEST ANSWER IS

have a great summer

1.01 x 10^-3 moles

**Answer:**

0.278 mol

**Explanation:**

**Step 1: Given and required data**

Mass of acetic acid (m): 16.7 g

Chemical formula of acetic acid: CH₃COOH (C₂H₄O₂)

**Step 2: Calculate the molar mass (M) of acetic acid**

We will use the following expression.

M(C₂H₄O₂) = 2 × M(C) + 4 × M(H) + 2 × M(O)

M(C₂H₄O₂) = 2 × 12.01 g/mol + 4 × 1.01 g/mol + 2 × 16.00 g/mol = 60.06 g/mol

**Step 3: Calculate the number of moles (n) of acetic acid**

We will use the following expression.

n = m/M

n = 16.7 g/(60.06 g/mol) = 0.278 mol

**Answer:**

** 2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O**

**α =2**

**β = 19**

**γ = 12**

**δ = 14**

**53.2moles of O₂**

**Explanation:**

Proper equation of the reaction:

αC₆H₁₄ + βO₂ → γCO₂ + δH₂O

This is a combustion reaction for a hydrocarbon. For the combustion of a hydrocarbon, the combustion equation is given below:

CₓHₙ + (x + )O₂ → xCO₂ + H₂O

From the given combustion equation, x = 6 and n = 14

Therefore:

β = x + = 6 + = 6 + 3.5 = 9

γ = 6

δ = = = 7

The complete reaction equation is therefore given as:

C₆H₁₄ + 9O₂ → 6CO₂ + 7H₂O

To express as whole number integers, we multiply the coefficients through by 2:

** 2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O**

**Problem 2**

** From the reaction: **

** 2 moles of hexane are required to completely react with 19 moles of O₂**

**∴ 5.6 moles of hexane would react with k moles of O₂**

This gives: 5.6 x 19 = 2k

k =

k =** 53.2moles of O₂**