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Novosadov [1.4K]
3 years ago
5

If a gas has a volume of 2 L at a pressure of 325 kPa, what volume will it have when the pressure is increased to 650 kPa? Assum

e the temperature and number of particles are constant.
Chemistry
1 answer:
Nastasia [14]3 years ago
7 0
P1v1=p2v2
v2=p1v1/p2
v2=325x2/650 = 1 L :)
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curie, <span>one basic unit are used for measuring radiation. One is the </span>curie,<span> named after Marie and Pierre Curie. </span>
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most radioactive atoms “spontaneously” decay to a more stable form. where does the “additional” energy go once the atoms have de
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When most radioactive atoms “spontaneously” decay to a more stable form the “additional” energy is converted to radiation with the emission of radioactive particles.

<h3>What is radioactivity?</h3>

Radioactivity is the spontaneous decay of the nucleus of an atom with the emission of radiation and nuclear particles.

Elements that spontaneously decay are called radioactive elements.

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1 year ago
Find the ph of a 0.250 m solution of nac2h3o2. (the ka value of hc2h3o2 is 1.80×10−5). express your answer numerically to four s
Slav-nsk [51]
Answer is: pH value of solution of NaC₂H₃O₂ is 9.07.
Chemical reaction: C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻.
Ka(HC₂H₃O₂) = 1,8·10⁻⁵.<span>
Ka · Kb = Kw.
</span>1,8·10⁻⁵ mol/dm³ · Kb = 1·10⁻¹⁴ mol²/dm⁶; the ionic product of water at 25°C.<span>
Kb(</span>C₂H₃O₂⁻) = 1·10⁻¹⁴ mol²/dm⁶ ÷ 1,8·10⁻⁵ mol/dm³.<span>
Kb(</span>C₂H₃O₂⁻) = 5,56·10⁻¹⁰ mol/dm³.
c(C₂H₃O₂⁻) = 0,25 M.
[OH⁻] = [HC₂H₃O₂] = x.
[C₂H₃O₂⁻] = 0,25 M - x.
Kb = [OH⁻] · [HC₂H₃O₂] / [C₂H₃O₂⁻].
5,56·10⁻¹⁰ = x² / (0,25 M -x).
Solve quadratic equation: x = [OH⁻] = 0,0000118 M.
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A 2.50 g sample of solid sodium hydroxide is added to 55.0 mL of 25 °C water in a foam cup (insulated from the environment) and
zlopas [31]

Answer:

37.1°C.

Explanation:

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<em>∵ ΔHsoln = Q/n</em>

no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.

<em>The negative sign of ΔHsoln indicates that the reaction is exothermic.</em>

∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.

  • We can use the relation:

Q = m.c.ΔT,

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m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).

c is the specific heat capacity of water (c = 4.18 J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).

∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)

∴ (final temperature - 25°C) = (2781.87 J)/(55.0 g)(4.18 J/g.°C) = 12.1.

<em>∴ final temperature = 25°C + 12.1 = 37.1°C.</em>

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