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Indian Navy AA (Artificer Apprentice) States of Matter — practice questions

Explore 56 free States of Matter questions in Indian Navy AA (Artificer Apprentice) Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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States of Matter questions and exercises with answers

Showing questions 1–30 of 56.

Boyle's law for a fixed amount of gas at constant temperature states that:Read the answer and worked solution →The hydrogen spectral lines that lie in the visible region of the electromagnetic spectrum belong to the:Read the answer and worked solution →The bond order of the oxygen molecule O₂ (by molecular orbital theory) isRead the answer and worked solution →CLASS10 · Chemistry · Periodic Table Trends — Gap FillRead the answer and worked solution →Sodium (atomic number 11) belongs to which group and period?Read the answer and worked solution →Which of the following elements will form an acidic oxide?Read the answer and worked solution →Order the atomic models as they were proposedRead the answer and worked solution →Match each scientist to their modelRead the answer and worked solution →Match each model to what it could NOT explainRead the answer and worked solution →True or false: Rutherford's modelRead the answer and worked solution →True or false: Bohr's modelRead the answer and worked solution →Complete the reason Bohr's model failsRead the answer and worked solution →Which model does this belong to?Read the answer and worked solution →Match each quantum number to what it is calledRead the answer and worked solution →True or false: the quantum mechanical modelRead the answer and worked solution →Order the structure: from atom down to electronRead the answer and worked solution →Complete the (n + l) ruleRead the answer and worked solution →True or false: orbital energiesRead the answer and worked solution →Match each rule to what it forbids or requiresRead the answer and worked solution →Tap every statement Hund's rule requiresRead the answer and worked solution →The fundamental particlesRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →Which principle is being described?Read the answer and worked solution →Flip and pair: symbol with meaningRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Tap every statement true of the Schrodinger modelRead the answer and worked solution →Reading the models backRead the answer and worked solution →Assertion and reason: filling orbitalsRead the answer and worked solution →

Worked practice questions

Boyle's law for a fixed amount of gas at constant temperature states that:

  1. Pressure is directly proportional to volume
  2. $P^2 V = $ constant
  3. Pressure is inversely proportional to volume
  4. Volume is independent of pressure
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Correct answer: C. Pressure is inversely proportional to volume

Boyle's law: at fixed $T$ and $n$, $PV = $ constant, so $P \propto 1/V$. Squeeze a gas into half the volume at constant temperature and its pressure doubles. The $P$-$V$ curve at fixed $T$ is a hyperbola called an isotherm. Option A is the opposite (direct proportion), which would mean compressing a gas reduces its pressure — physically absurd.

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The hydrogen spectral lines that lie in the visible region of the electromagnetic spectrum belong to the:

  1. Lyman series
  2. Paschen series
  3. Pfund series
  4. Balmer series
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Correct answer: D. Balmer series

Balmer series corresponds to transitions terminating in $n_1 = 2$. The first four lines ($n_2 = 3, 4, 5, 6$ to $n_1 = 2$) lie at approximately $656\,\text{nm}$ (H$_\alpha$, red), $486\,\text{nm}$ (H$_\beta$, blue-green), $434\,\text{nm}$ (H$_\gamma$, blue) and $410\,\text{nm}$ (H$_\delta$, violet), all in the visible range.

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The bond order of the oxygen molecule O₂ (by molecular orbital theory) is

  1. 2
  2. 1
  3. 2.5
  4. 3
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Correct answer: A. 2

MO config gives bond order = (bonding − antibonding)/2 = (8−4)/2 = 2, consistent with the O=O double bond.

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More Chemistry topics

Chemical EquilibriumAlcohols Phenols EthersAlkenesOrganic ChemistryPolymersAtoms and MoleculesClassification of ElementsS Block Elements