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3 years ago

How should the enthalpy of an intermediate step be manipulated when used to produce an overall chemical equation?

1 answer:

8 0

The enthalpy of an intermediate step should be manipulated when used to produce an overall equation by using the Hess's law. You could multiply the enthalpy by -1 if this equation is reversed in theory.

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What kind of chemical reaction does the chemical equation sodium + chlorine → sodium chloride represent?

Vinil7 [7]

D. Synthesis cause A + B --> AB Equation: 2Na + Cl2 --> 2NaCl

7 0

3 years ago

Read 2 more answers

What is the volume of 3.00 M sulfuric aid that contain 9.809 g of H2SO4 (98.09g/mol)

Slav-nsk [51]

Given :

Molarity of sulfuric acid solution is 3.0 M.

Amount of sulfuric acid present in solution is 9.809 g.

To Find :

The volume of solution.

Solution :

We know, molarity is given by :

$Molarity = \dfrac{number \ of \ moles \times 1000}{Volume\ ( ml )}\\\\M = \dfrac{w \times 1000}{M.M \times V}\\\\3 = \dfrac{9.809\times 1000}{98.09 \times V}\\\\V = \dfrac{1000}{10\times 3}\ ml\\\\V = 33.33 \ ml$

Therefore, volume required is 33.33 ml .

5 0

3 years ago

g If 50.0 mL of a 0.75 M acetic acid solution is titrated with 1.0 M sodium hydroxide, what is the pH after 10.0 mL of NaOH have

V125BC [204]

Answer:

pH = 2.66

Explanation:

Acetic Acid + NaOH → Sodium Acetate + H₂O

First we calculate the number of moles of each reactant, using the given volumes and concentrations:

0.75 M Acetic acid * 50.0 mL = 37.5 mmol acetic acid

1.0 M NaOH * 10.0 mL = 10 mmol NaOH

We calculate how many acetic acid moles remain after the reaction:

37.5 mmol - 10 mmol = 27.5 mmol acetic acid

We now calculate the molar concentration of acetic acid after the reaction:

27.5 mmol / (50.0 mL + 10.0 mL) = 0.458 M

Then we calculate [H⁺], using the following formula for weak acid solutions:

[H⁺] = $\sqrt{C*Ka}=\sqrt{0.458M*1.76x10^{-5}}$

[H⁺] = 0.0028

Finally we calculate the pH:

pH = -log[H⁺]

pH = 2.66

8 0

3 years ago

Isoelectric point. The isoelectric point (pI) is the pH at which a molecule has no net charge. The amino acid glycine has two io

timama [110]

Answer:

pI = 6.16

Explanation:

The pI is given by the average of the pKas that are involved. In this case,

Pka of carboxylic acid was given as 2.72 and that of the Amino group was given as 9.60. the average would then be ½(2.72+9.60)

= 6.16

6 0

3 years ago

0.25 mol of nitric acid is used to make a 0.10 M nitric acid solution. How many grams of solute was used?

Yanka [14]

Answer:

15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution

Explanation:

Step 1: Data given

Nitric acid = HNO3

Molar mass of H = 1.01 g/mol

Molar mass of N = 14.0 g/mol

Molar mass O = 16.0 g/mol

Number of moles nitric acid (HNO3) = 0.25 moles

Molairty = 0.10 M

Step 2: Calculate molar mass of nitric acid

Molar mass HNO3 = Molar mass H + molar mass N + molar mass (3*O)

Molar mass HNO3 = 1.01 + 14.0 + 3*16.0

Molar mass HNO3 = 63.01 g/mol

Step 3: Calculate mass of solute use

Mass HNO3 = moles HNO3 * molar mass HNO3

Mass HNO3 = 0.25 moles * 63.01 g/mol

Mass HNO3 = 15.75 grams

15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution

7 0

3 years ago