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Elanso [62]
1 year ago
12

How many carbon monoxide molecules are in 0.75 moles of carbon monoxide?

Chemistry
2 answers:
Flauer [41]1 year ago
6 0

Answer:

0.75 ( 6.02 × 10²³) = 4.515 x 10²³

Explanation:

0.75 ( 6.02 × 10²³) = 4.515 x 10²³

Hunter-Best [27]1 year ago
3 0

Answer:

four 0.5 times 10 to the 23rd Adams of Neptune

Explanation:

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a substance did not change its chemical nature in a reaction. which most likely describes the reaction
Mkey [24]
This substance most likely is an inert. It is a substance that is not chemically reactive. It does not change its chemical nature in a reaction. It does not <span>easily react with other chemicals. Most of the group 8 gases in the periodic table are classified as inert, due to their having full outer electron shells. </span>
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3 years ago
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Calculate the concentrations of all species present in a 0.26 M solution of ethylammonium chloride (C2H5NH3Cl).
Alina [70]

Answer:

0.00000223

Explanation:

pKa for C2H5NH3+ = 10.7

pKw = 14.0

pKa + pKb = pKw

10.7 + pKb = 14.0

pKb = 14.0 - 10.7

pKb = 3.30

C2H5NH3Cl is a salt of ethylamine and HCl so it will dissolve in water to produce  C2H5NH3^+ + Cl^-

The base hydrolysis reaction:  C2H5NH3^+(aq) + H2O(l) <=> C2H5NH2(aq) + H3O^+(aq)

This reaction is described by Kb.

Kb = [C2H5NH2][H3O^+]/[C2H5NH3^+]

Let [C2H5NH2] = [H3O^+] = x,

so [C2H5NH3^+] = 0.26 - x

Kb = x^2/(0.26 - x) = 2.00 x 10^-11  

Let's solve for x. In this equation,  It is possible to solve without the use quadratic equation. So we can assume that 0.26 - x  is approximately equal to 0.26.  We won't know until we do the calculation.

We get:  x^2 + 2.00 x 10^-11x - 4.99 x 10^-12 = 0

With the use of a quadratic calculator.

x = 2.23 x 10^-6 M = [C2H5NH2] = [H3O^+]

0.26 - x  is just 0.26 M in this problem because 2.23 x 10^-6 M is insignificant.

[C2H5NH3^+] = 0.26 M = [Cl^-]

NOTE:

pH = -log [H3O^+] = -log(2.23 x 10^-6) = 5.65

Ka is the acid dissociation constant

Kb is the base dissociation constant

5 0
3 years ago
Which of the following describes the physical state of the layers of the Earth from the outside to the center?
Phantasy [73]
Solid , solid, partially melted, liquid so the answer is D
7 0
3 years ago
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You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.1950 M NaOH sol
oee [108]

Answer:

0.2788 M

1.674 %(m/V)

Explanation:

Step 1: Write the balanced equation

NaOH + CH₃COOH → CH₃COONa + H₂O

Step 2: Calculate the reacting moles of NaOH

0.03575 L \times \frac{0.1950mol}{L} = 6.971 \times 10^{-3} mol

Step 3: Calculate the reacting moles of CH₃COOH

The molar ratio of NaOH to CH₃COOH is 1:1.

6.971 \times 10^{-3} molNaOH \times \frac{1molCH_3COOH}{1molNaOH} = 6.971 \times 10^{-3} molCH_3COOH

Step 4: Calculate the molarity of the acetic acid solution

M = \frac{6.971 \times 10^{-3} mol}{0.02500L} =0.2788 M

Step 5: Calculate the mass of acetic acid

The molar mass of acetic acid is 60.05 g/mol.

6.971 \times 10^{-3} mol \times \frac{60.05g}{mol} =0.4186 g

Step 6: Calculate the percentage of acetic acid in the solution

\frac{0.4186g}{25.00mL}  \times 100\% = 1.674 \%(m/V)

6 0
3 years ago
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An analytical chemist is titrating 242.5mL of a 1.200M solution of hydrazoic acid HN3 with a 0.3400M solution of NaOH . The pKa
fgiga [73]

<u>Answer:</u> The pH of the solution is 12.61

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}       ......(1)

  • <u>For hydrazoic acid:</u>

Molarity of hydrazoic acid solution = 1.200 M

Volume of solution = 242.5 mL

Putting values in equation 1, we get:

1.200M=\frac{\text{Moles of hydrazoic acid}\times 1000}{242.5mL}\\\\\text{Moles of hydrazoic acid}=0.291mol

  • <u>For NaOH:</u>

Molarity of NaOH solution = 0.3400 M

Volume of solution = 1006 mL

Putting values in equation 1, we get:

0.3400M=\frac{\text{Moles of NaOH}\times 1000}{1006mL}\\\\\text{Moles of NaOH}=0.342mol

The chemical reaction for hydrazoic acid and NaOH follows the equation:

                   HN_3+NaOH\rightarrow NaN_3+H_2O

<u>Initial:</u>           0.291        0.342

<u>Final:</u>                  0          0.051                 0.291      0.291

Volume of solution = 242.5 + 1006 = 1248.5 mL = 1.2485 L    (Conversion factor:  1 L = 1000 mL)

  • <u>For NaOH left:</u>

Left moles of NaOH = 0.051 moles

Volume of the solution = 1.2485 L

Putting values in equation 1, we get:

\text{Molarity of NaOH}=\frac{0.051mol}{1.2485L}=0.0408M

1 mole of NaOH produces 1 mole of sodium ions and 1 mole of hydroxide ions

To calculate pOH of the solution, we use the equation:

pOH=-\log[OH^-]

We are given:

[OH^-]=0.0408M

Putting values in above equation, we get:

pOH=-\log(0.0408)\\\\pOH=1.39

To calculate pH of the solution, we use the equation:

pH+pOH=14\\\\pH=14-1.39=12.61

Hence, the pH of the solution is 12.61

8 0
3 years ago
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