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
allsm [11]
3 years ago
9

Neon gas has a volume of 261 ml and a pressure of 108.5 kPa. What volume will the gas occupy at 96.5 kPa if the temperature rema

ins constant?
Chemistry
1 answer:
Tatiana [17]3 years ago
5 0

Answer:

\large \boxed{\text{293 mL}}

Explanation:

The temperature and amount of gas are constant, so we can use Boyle’s Law.

p_{1}V_{1} = p_{2}V_{2}

Data:

\begin{array}{rcrrcl}p_{1}& =& \text{108.5 kPa}\qquad & V_{1} &= & \text{261 mL} \\p_{2}& =& \text{96.5 kPa}\qquad & V_{2} &= & ?\\\end{array}

Calculations:  

\begin{array}{rcl}108.5 \times 261 & =& 96.5V_{2}\\28320 & = & 96.5V_{2}\\V_{2} & = &\dfrac{28320}{96.5}\\\\& = &\textbf{293 mL}\\\end{array}\\\text{The volume of the gas is } \large \boxed{\textbf{293 mL}}

You might be interested in
Why should we try to use less oil, natural gas and electricity?
oksano4ka [1.4K]
Because they aren’t sustainable and they contribute to global warming, I think.
6 0
3 years ago
Read 2 more answers
Н - О - Н<br> I have to find the Molecular formula
Vladimir [108]
Answer
H2O is the answer
5 0
2 years ago
Describe the relationship between kinetic energy and mass
zhannawk [14.2K]

Answer:

Kinetic energy has a direct relationship with mass, meaning that as mass increases so does the Kinetic Energy of an object. ... Objects with greater mass can have more kinetic energy even if they are moving more slowly, and objects moving at much greater speeds can have more kinetic energy even if they have less mass

8 0
3 years ago
Read 2 more answers
Briefly discuss the 3 most important properties of metals
Elan Coil [88]

Answer: Luster: Metals are shiny when cut, scratched, or polished.

Malleability: Metals are strong but malleable, which means that they can be easily bent or shaped.

Conductivity: Metals are excellent conductors of electricity and heat.

Explanation: Hope this helped!

7 0
2 years ago
A 20.0 mL 0.100 M solution of lactic acid is titrated with 0.100 M NaOH.
yan [13]

Answer:

(a) See explanation below

(b) 0.002 mol

(c) (i) pH = 2.4

(ii) pH = 3.4

(iii) pH = 3.9

(iv) pH = 8.3

(v) pH = 12.0

Explanation:

(a) A buffer solution exits after addition of 5 mL of NaOH  since after reaction we will have  both the conjugate base lactate anion and unreacted weak  lactic acid present in solution.

Lets call lactic acid HA, and A⁻ the lactate conjugate base. The reaction is:

HA + NaOH ⇒ A⁻ + H₂O

Some unreacted HA will remain in solution, and since HA is a weak acid , we will have the followin equilibrium:

HA  + H₂O ⇆ H₃O⁺ + A⁻

Since we are going to have unreacted acid, and some conjugate base, the buffer has the capacity of maintaining the pH in a narrow range if we add acid or base within certain limits.

An added acid will be consumed by the conjugate base A⁻ , thus keeping the pH more or less equal:

A⁻ + H⁺ ⇄ HA

On the contrary, if we add extra base it will be consumed by the unreacted lactic acid, again maintaining the pH more or less constant.

H₃O⁺ + B ⇆ BH⁺

b) Again letting HA stand for lactic acid:

mol HA =  (20.0 mL x  1 L/1000 mL) x 0.100 mol/L = 0.002 mol

c)

i) After 0.00 mL of NaOH have been added

In this case we just have to determine the pH of a weak acid, and we know for a monopric acid:

pH = - log [H₃O⁺] where  [H₃O⁺] = √( Ka [HA])

Ka for lactic acid = 1.4 x 10⁻⁴  ( from reference tables)

[H₃O⁺] = √( Ka [HA]) = √(1.4 x 10⁻⁴ x 0.100) = 3.7 x 10⁻³

pH = - log(3.7 x 10⁻³) = 2.4

ii) After 5.00 mL of NaOH have been added ( 5x 10⁻³ L x 0.1 = 0.005 mol NaOH)

Now we have a buffer solution and must use the Henderson-Hasselbach equation.

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.0005                0

after rxn    0.002-0.0005                  0                  0.0005

                        0.0015

Using Henderson-Hasselbach equation :

pH = pKa + log [A⁻]/[HA]

pKa HA = -log (1.4 x 10⁻⁴) = 3.85

pH = 3.85 + log(0.0005/0.0015)

pH = 3.4

iii) After 10.0 mL of NaOH have been ( 0.010 L x 0.1 mol/L = 0.001 mol)

                             HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.001               0

after rxn        0.002-0.001                  0                  0.001

                        0.001

pH = 3.85 + log(0.001/0.001)  = 3.85

iv) After 20.0 mL of NaOH have been added ( 0.002 mol )

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.002                 0

after rxn                 0                         0                   0.002

We are at the neutralization point and  we do not have a buffer anymore, instead we just have  a weak base A⁻ to which we can determine its pOH as follows:

pOH = √Kb x [A⁻]

We need to determine the concentration of the weak base which is the mol per volume in liters.

At this stage of the titration we added 20 mL of lactic acid and 20 mL of NaOH, hence the volume of solution is 40 mL (0.04 L).

The molarity of A⁻ is then

[A⁻] = 0.002 mol / 0.04 L = 0.05 M

Kb is equal to

Ka x Kb = Kw ⇒ Kb = 10⁻¹⁴/ 1.4 x 10⁻⁴ = 7.1 x 10⁻¹¹

pOH is then:

[OH⁻] = √Kb x [A⁻]  = √( 7.1 x 10⁻¹¹ x 0.05) = 1.88 x 10⁻⁶

pOH = - log (  1.88 x 10⁻⁶ ) = 5.7

pH = 14 - pOH = 14 - 5.7 = 8.3

v) After 25.0 mL of NaOH have been added (

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn           0.002                  0.0025              0

after rxn                0                         0.0005              0.0005

Now here what we have is  the strong base sodium hydroxide and A⁻ but the strong base NaOH will predominate and drive the pH over the weak base A⁻.

So we treat this part as the determination of the pH of a strong base.

V= (20 mL + 25 mL) x 1 L /1000 mL = 0.045 L

[OH⁻] = 0.0005 mol / 0.045 L = 0.011 M

pOH = - log (0.011) = 2

pH = 14 - 1.95 = 12

7 0
2 years ago
Other questions:
  • What is a isotope ?
    15·2 answers
  • how is the concept of effective nuclear charge used to simplify the numerous electron-electron repulsions in many-electron atom
    15·1 answer
  • Ammonium perchlorate (NH4ClO4) is the solid rocket fuel used by the U.S. Space Shuttle. It reacts with itself to produce nitroge
    12·1 answer
  • The compound F2O has two lone pairs on the central oxygen atom. What's the structure of an F2O molecule? Is this compound polar
    6·1 answer
  • Which formula represents a molecular compound?<br> (1) HI (3) KCl<br> (2) KI (4) LiCl
    9·2 answers
  • 400ml of 2.5M can be deluded to make how many of 2.0 HCI
    12·1 answer
  • El nivel simbólico representa tanto a lo macroscópico como a lo nanoscópico.
    10·1 answer
  • How many atoms are in 5Cao4?
    12·1 answer
  • The main function of the respiratory system is to –
    13·2 answers
  • Dehydration of tertiary alcohols occurs by what mechanism?
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!