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
vova2212 [387]
3 years ago
7

Under which circumstance might a gas decrease in volume when heated? A)The gas is held constant at STP. B)The gas remains under

uniform pressure. C)tHE GAS IS PLACE UNDER INCREASING PRESSURE. D)The gas undergoes a decrease in pressure
Chemistry
1 answer:
Tcecarenko [31]3 years ago
3 0

Answer is option C. The gas is placed under increased pressure.

Explanation:

According to Gas laws,

If the temperature to gas increases then the molecules of gas gets additional energy and moves with more spped than earlier which gives more more colliding to the walls of container. This increases pressure.

The volume and gas in a container are directly proportional to each other.

This means if the volume gets decreased then the collision of gas molecules to the walls increases which results increase in pressure.

Therefore, volume is inversely proportional to pressure.

You might be interested in
Identify the standard metal cations in the active components of soaps
rjkz [21]
Answer is: standard metal cations are sodium cation (Na⁺), potassium cation (K⁺), magnesium cation (Mg²⁺) and calcium cation (Ca²⁺).
Soap<span> is a </span>salt<span> of a </span>fatty acid. When soap have sodium and potassim cations, that is toilet soap and when soap have magnesium and calcium cations, that soap is called <span>metallic soap.</span>
8 0
3 years ago
A 50.0 mL solution of 0.129 M KOH is titrated with 0.258 M HCl. Calculate the pH of the solution after the addition of each of t
kobusy [5.1K]

Answer:

A- pH = 13.12

B- pH = 12.91

C- pH = 12.71

D- pH = 12.43

E- pH = 11.55

F- pH = 7

G- pH = 2.46

H- pH = 1.88

Explanation:

This is a titration of a strong base with a strong acid. The neutralization reaction is: KOH (aq) + HCl (aq) →  H₂O(l) + KCl(aq)

Our pH at the equivalence point is 7, because we have made a neutral salt.

To determine the volume at that point we state the formula for titration:

mmoles of base = mmoles of acid

Volume of base  . M of base = Volume of acid . M of acid

50mL . 0.129M = 0.258 M . Volume of acid

Volume of acid = (50mL . 0.129M) / 0.258 M →  25 mL (Point <u>F</u>)

When we add 25 mL of HCl, our pH will be 7.

A- At 0 mL of acid, we only have base.

KOH → K⁺ + OH⁻

[OH⁻] = 0.129 M

To make more easy the operations we will use, mmol.

mol . 1000 = mmoles → mmoles / mL = M

- log 0.129 = 0.889

14 - 0.889 = 13.12

B-  In this case we are adding, (7 mL . 0.258M) = 1.81 mmoles of H⁺

Initially we have  0.129 M . 50 mL = 6.45 mmoles of OH⁻

1.81 mmoles of H⁺ will neutralize, the 6.45 mmoles of OH⁻ so:

6.45 mmol - 1.81 = 4.64 mmoles of OH⁻

This mmoles of OH⁻ are not at 50 mL anymore, because our volume has changed. (Now, we have 50 mL of base + 7 mL of acid) = 57 mL of total volume.

[OH⁻] = 4.64 mmoles / 57 mL = 0.0815 M

- log 0.0815 M = 1.09 → pOH

pH = 14 - pOH → 14 - 1.09 = 12.91

C- In this case we add (12.5 mL . 0.258M) = 3.22 mmoles of H⁺

<em>Our initial mmoles of OH⁻ would not change through all the titration. </em>

Then 6.45 mmoles of OH⁻ are neutralized by 3.22 mmoles of H⁺.

6.45 mmoles of OH⁻ - 3.22 mmoles of H⁺ = 3.23 mmoles of OH⁻

Total volume is: 50 mL of base + 12.5 mL = 62.5 mL

[OH⁻] = 3.23 mmol / 62.5 mL = 0.0517 M

- log  0.0517 = 1.29 → pOH

14 - 1.11 = 12.71

D- We add (18 mL . 0.258M) = 4.64 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 4.64 mmoles of H⁺.

6.45 mmoles of OH⁻ - 4.64 mmoles of H⁺ = 1.81 mmoles of OH⁻

Total volume is: 50 mL of base + 18 mL = 68 mL

[OH⁻] = 1.81 mmol / 68 mL = 0.0265 M

- log  0.0265 = 1.57 → pOH

14 - 1.57 = 12.43

E- We add (24 mL . 0.258M) = 6.19 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 6.19 mmoles of H⁺.

6.45 mmoles of OH⁻ - 6.19 mmoles of H⁺ = 0.26 mmoles of OH⁻

Total volume is: 50 mL of base + 24 mL = 74 mL

[OH⁻] = 0.26 mmol / 74 mL = 3.51×10⁻³ M

- log  3.51×10⁻³  = 2.45 → pOH

14 - 2.45 = 11.55

F- This the equivalence point.

mmoles of OH⁻ = mmoles of H⁺

We add (25 mL . 0.258M) = 6.45 mmoles of H⁺

All the OH⁻ are neutralized.

OH⁻  +  H⁺  ⇄   H₂O              Kw

[OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷  →  pOH = 7

pH → 14 - 7 = 7

G- In this case we have an excess of H⁻

We add (26 mL . 0.258M ) = 6.71 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

6.71 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 0.26 mmoles of H⁺

[H⁺] = 0.26 mmol / Total volume

Total volume is: 50 mL + 26 mL → 76 mL

[H⁺] = 0.26 mmol / 76 mL → 3.42×10⁻³ M

- log 3.42×10⁻³ = 2.46 → pH

H- Now we add (29 mL . 0.258M) = 7.48 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

7.48 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 1.03 mmoles of protons

Total volume is 50 mL + 29 mL = 79 mL

[H⁺] = 1.03 mmol / 79 mL → 0.0130 M

- log 0.0130 = 1.88 → pH

After equivalence point, pH will be totally acid, because we always have an excess of protons. Before the equivalence point, pH is basic, because we still have OH⁻ and these hydroxides, will be neutralized through the titration, as we add acid.

5 0
3 years ago
What is the correct formula to determine density?
wlad13 [49]
D=M/V I hope this helps
7 0
3 years ago
Read 2 more answers
What is the scientific notation of 68000
Kaylis [27]
68000 = 6.8 * 10000 = 6.8 * 10^4  

hope this helps? c;
3 0
3 years ago
Can somebody check and tell me the answer of this question pls?
andriy [413]
The answer is A combustion of allâmes
3 0
3 years ago
Other questions:
  • If a sample of gold is a cube what is the length of each edge in centimeters
    13·1 answer
  • At room temperature, 20ÁC, water (H2O) is a liquid and oxygen (O2) is a gas. Based on kinetic molecular theory, which answer bes
    6·1 answer
  • Given 65 g of C3 H8 and 65 g of O2, what is the theoretical yield of CO2 in grams
    6·1 answer
  • 2' A mixture containing 2.75 gof ammonium chloride (NH4cl) in 5.0 g of water was heated to dissolve the solid and then allowed t
    6·1 answer
  • Doctors sometimes use computer models to track the spread of an infectious disease . How might this help them better understand
    11·1 answer
  • A white substance melts with some decomposition at 730oC. As a solid it does not conduct electricity, but it dissolves in water
    14·1 answer
  • Find the volume in m3 of 52.6 lbm of iron:
    13·1 answer
  • HELP Which type of circuit would be best to use for lights used for decorations?
    13·1 answer
  • Please please !!! help me I’ve been doing this for 2 hours
    13·1 answer
  • A balanced chemical equation has the same number and type of atoms in the reactants as compared to the products. An unbalanced c
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!