1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
inysia [295]
2 years ago
9

HELP!!!!!

Chemistry
2 answers:
faltersainse [42]2 years ago
5 0

Answer:

5.17.

Explanation:

<em>∵ [H₃O⁺][OH⁻] = 10⁻¹⁴. </em>

[OH⁻] = 1.5  x 10⁻⁹ M.

∴ [H₃O⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(1.5  x 10⁻⁹ M) = 6.66 × 10⁻⁶ M.

∵ pH = - log[H₃O⁺]

<em>∴ pH = - log(6.66 × 10⁻⁶ M) = 5.17. </em>

Elena-2011 [213]2 years ago
4 0

Answer:

5.18

Explanation:

I hope this helps ;)

You might be interested in
Using this illustration, what is the chemical formula? Picture and possible answers are below.
qwelly [4]
NH3. 1 nitrogen and 3 hydrogen
4 0
3 years ago
Read 2 more answers
The estimated heat of vaporization of diethyl ether using the Chen's rule is A. 29.7 KJ/mol B. 33.5 KJ/mol C. 26.4 KJ/mol D. 36.
Brums [2.3K]

Answer:

C. 26.4 kJ/mol

Explanation:

The Chen's rule for the calculation of heat of vaporization is shown below:

\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]

Where,

\Delta H_v is the Heat of vaoprization (J/mol)

T_b is the normal boiling point of the gas (K)

T_c is the Critical temperature of the gas (K)

P_c is the Critical pressure of the gas (bar)

R is the gas constant (8.314 J/Kmol)

For diethyl ether:

T_b=307.4\ K

T_c=466.7\ K

P_c=36.4\ bar

Applying the above equation to find heat of vaporization as:

\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]

\Delta H_v=26400 J/mol

The conversion of J into kJ is shown below:

1 J = 10⁻³ kJ

Thus,

\Delta H_v=26.4 kJ/mol

<u>Option C is correct</u>

6 0
3 years ago
Which material is likely to slow the flow of electric charges the most? Explain
vesna_86 [32]

Answer:

The material that would most likely slow the flow of electricity is plastic.

Explanation:

Plastic is a conductor. A conductor is something that stop or slows down the flow of electricity.

4 0
2 years ago
How many grams of CO2 are in 3.6 mol of CO2?
hammer [34]

There are 158.4 grams of CO2 in 3.6 mol of CO2.

<h3>HOW TO CALCULATE MASS?</h3>

The mass of a substance can be calculated by multiplying the number of moles of the substance by its molar mass. That is;

mass of CO2 = no. of moles × molar mass

According to this question, there are 3.6 moles of CO2.

mass of CO2 = 3.6 moles × 44g/mol

mass of CO2 = 158.4g.

Therefore, there are 158.4 grams of CO2 in 3.6 mol of CO2

Learn more about mass at: brainly.com/question/15959704

6 0
2 years ago
Read 2 more answers
In a exothermic reaction, is heat transfered to or from the surrounding
Monica [59]
In an exothermic reaction, heat is transferred to the surrounding.

Hope this helps, have a great day ahead!
7 0
3 years ago
Read 2 more answers
Other questions:
  • how would physical changes to Earth such as an iceberg or rising sea levels lead to physical changes in organisms?
    11·1 answer
  • What is the total mass of three gold bars that weigh 5543 mg, 23.45 mg, and 697.4 mg?
    6·1 answer
  • What is the enthalpy for reaction 1 reversed? reaction 1 reversed: 2CO2+3H2O→C2H5OH+3O2?
    11·2 answers
  • Suppose you set a glass of water in direct sunlight for 2 hours and measure its temperature every 10 min . What type of graph wo
    9·1 answer
  • Which of the following is an sxample of using physical capital to save time and money
    15·1 answer
  • I JUST NEED HELP!!!!
    15·1 answer
  • Which of the following does NOT show a chemical change of matter?
    7·1 answer
  • Potential Energy vs Height
    10·1 answer
  • Conservation Models Activity
    10·1 answer
  • Pt 2. Chem Reactions 50 PTS
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!