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GRADE 7 PRACTICE

Physics Problems: Structure of matter, Grade 7

Particles, diffusion, interaction and states of aggregation.

18 problems across 4 sub-topics
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Problem statements

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18 problems
Position in this set: 1 from 18

Problem #7

Particles and diffusion

What concept does the description correspond to: “A substance consists of tiny particles with gaps between them; the particles are always moving and interacting.»?

Answer options

  1. \text{Molecular structure}
  2. \text{Brownian motion}
  3. \text{Diffusion}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{A substance consists of tiny particles with gaps between them; the particles are always moving and interacting.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    The sizes of molecules are much smaller than usual bodies, so their properties are determined by observed effects and special instruments.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Molecular structure}

Answer\text{Concept} = \text{Molecular structure}

Position in this set: 2 from 18

Problem #8

Particles and diffusion

What concept does the description correspond to: “Random movement of small particles in a liquid or gas due to uneven impacts of the molecules of the medium.»?

Answer options

  1. \text{Brownian motion}
  2. \text{Molecular structure}
  3. \text{Diffusion}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{Random movement of small particles in a liquid or gas due to uneven impacts of the molecules of the medium.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    Pollen seen under a microscope changes direction without an external nudge, providing evidence of random thermal molecular motion.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Brownian motion}

Answer\text{Concept} = \text{Brownian motion}

Position in this set: 3 from 18

Problem #9

Structure of matter

Mixing of substances is observed in four identical vessels. Where, other things being equal, will diffusion go faster?

Answer options

  1. \text{B warmth gas}
  2. \text{B cold hard body}
  3. \text{B cold liquids}
  4. \text{the same in everyone environments}
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Given
\text{Terms}
\text{different state and temperature}
Find

\text{the greatest Diffusion rate}

  1. \text{gas: particles freer are moving}
  2. \text{heating: thermal movement more intense}\text{warm gas}

Answer\text{B warmth gas}

Position in this set: 4 from 18

Problem #10

Interaction between particles and states of matter

What concept does the description correspond to: “At short distances, particles of matter simultaneously exhibit attraction and resistance to excessive proximity.»?

Answer options

  1. \text{Interactions between molecules}
  2. \text{Solid state}
  3. \text{Liquid and gas}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{At short distances, particles of matter simultaneously exhibit attraction and resistance to excessive proximity.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    Attraction holds an object’s parts together; rapidly increasing repulsion prevents a solid from being compressed to zero volume.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Interactions between molecules}

Answer\text{Concept} = \text{Interactions between molecules}

Position in this set: 5 from 18

Problem #11

Interaction between particles and states of matter

What concept does the description correspond to: “A state in which particles vibrate around stable positions and the object keeps its shape and volume.»?

Answer options

  1. \text{Solid state}
  2. \text{Interactions between molecules}
  3. \text{Liquid and gas}
  4. \text{Physical phenomenon}
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Given
\text{Description}
\text{A state in which particles vibrate around stable positions and the object keeps its shape and volume.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    A crystal resists changing shape because its particles are more connected and ordered than in a liquid.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Solid state}

Answer\text{Concept} = \text{Solid state}

Position in this set: 6 from 18

Problem #12

Aggregate states

Which property best distinguishes a gas from a liquid in a closed container?

Answer options

  1. \text{gas takes all affordable volume}
  2. \text{gas saves own form}
  3. \text{gas saves own volume}
  4. \text{Particles gas motionless}
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Given
\text{Vessel}
\text{closed}
Find

\text{Property gas}

  1. \text{Particles gas far away friend from friend and moving free}
  2. \text{Therefore gas expanding}\text{fills in vessel}

Answer\text{takes all affordable volume}

Position in this set: 7 from 18

Problem #70

Particles and diffusion

For dissemination dye observed 7 min. Express duration experience B seconds.

Answer options

  1. 420\,\text{с}
  2. 840\,\text{с}
  3. 210\,\text{с}
  4. 1260\,\text{с}
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Given
t
7\,\text{мин}
Find

T

  1. T=60tT=7 \cdot 60T=420\,\text{с}

    It is convenient to compare durations in the same units.

Answer420\,\text{с}

Position in this set: 8 from 18

Problem #71

Particles and diffusion

For dissemination dye observed 9 min. Express duration experience B seconds.

Answer options

  1. 1080\,\text{с}
  2. 540\,\text{с}
  3. 270\,\text{с}
  4. 1620\,\text{с}
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Given
t
9\,\text{мин}
Find

T

  1. T=60tT=9 \cdot 60T=540\,\text{с}

    It is convenient to compare durations in the same units.

Answer540\,\text{с}

Position in this set: 9 from 18

Problem #72

Particles and diffusion

B cold water dye reached tags through 6 min, B warm through 90 s. In how much times first Time interval more second? Tags are on the same distance.

Answer options

  1. 8\,\text{раз}
  2. 4\,\text{раз}
  3. 2\,\text{раз}
  4. 12\,\text{раз}
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Given
t_1
6\,\text{мин}
t_2
90\,\text{с}
Find

r

  1. T=60t_1T=6 \cdot 60T=360\,\text{с}

    Convert minutes to seconds.

  2. r=\frac{T}{t_2}r=\frac{360}{90}r=4\,\text{раз}

    We compare the measured times. This is not the law of linear diffusion.

Answer4\,\text{раз}

Position in this set: 10 from 18

Problem #73

Particles and diffusion

B cold water dye reached tags through 8 min, B warm through 120 s. In how much times first Time interval more second? Tags are on the same distance.

Answer options

  1. 8\,\text{раз}
  2. 2\,\text{раз}
  3. 4\,\text{раз}
  4. 12\,\text{раз}
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Given
t_1
8\,\text{мин}
t_2
120\,\text{с}
Find

r

  1. T=60t_1T=8 \cdot 60T=480\,\text{с}

    Convert minutes to seconds.

  2. r=\frac{T}{t_2}r=\frac{480}{120}r=4\,\text{раз}

    We compare the measured times. This is not the law of linear diffusion.

Answer4\,\text{раз}

Position in this set: 11 from 18

Problem #74

Particles and diffusion

B cold water two identical experience occupied 280 and 320 with; B warm 90 and 110 s. In how much times Average time B cold water more average time B warm?

Answer options

  1. 6\,\text{раз}
  2. 1{,}5\,\text{раз}
  3. 3\,\text{раз}
  4. 9\,\text{раз}
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Given
a
280\,\text{с}
b
320\,\text{с}
c
90\,\text{с}
d
110\,\text{с}
Find

r

  1. T_1=\frac{a+b}{2}T_1=\frac{(280 + 320)}{2}T_1=300\,\text{с}

    We average repeated measurements in cold water.

  2. T_2=\frac{c+d}{2}T_2=\frac{(90 + 110)}{2}T_2=100\,\text{с}

    Separately, we average the measurements in warm water.

  3. r=\frac{T_1}{T_2}r=\frac{300}{100}r=3\,\text{раз}

    Temperature is a variable experimental condition; other conditions are the same.

Answer3\,\text{раз}

Position in this set: 12 from 18

Problem #75

Particles and diffusion

B cold water two identical experience occupied 460 and 500 with; B warm 150 and 170 s. In how much times Average time B cold water more average time B warm?

Answer options

  1. 6\,\text{раз}
  2. 1{,}5\,\text{раз}
  3. 9\,\text{раз}
  4. 3\,\text{раз}
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Given
a
460\,\text{с}
b
500\,\text{с}
c
150\,\text{с}
d
170\,\text{с}
Find

r

  1. T_1=\frac{a+b}{2}T_1=\frac{(460 + 500)}{2}T_1=480\,\text{с}

    We average repeated measurements in cold water.

  2. T_2=\frac{c+d}{2}T_2=\frac{(150 + 170)}{2}T_2=160\,\text{с}

    Separately, we average the measurements in warm water.

  3. r=\frac{T_1}{T_2}r=\frac{480}{160}r=3\,\text{раз}

    Temperature is a variable experimental condition; other conditions are the same.

Answer3\,\text{раз}

Position in this set: 13 from 18

Problem #76

Interaction between particles and states of matter

gas occupied 80 ml, A after compression 50 ml. On how much milliliters decreased him volume?

Answer options

  1. 60\,\text{мл}
  2. 30\,\text{мл}
  3. 15\,\text{мл}
  4. 90\,\text{мл}
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Given
V_1
80\,\text{мл}
V_2
50\,\text{мл}
Find

\Delta V

  1. \Delta V=V_1-V_2\Delta V=80 - 50\Delta V=30\,\text{мл}

    During compression, the distance between particles changes mainly.

Answer30\,\text{мл}

Position in this set: 14 from 18

Problem #77

Interaction between particles and states of matter

gas occupied 120 ml, A after compression 70 ml. On how much milliliters decreased him volume?

Answer options

  1. 100\,\text{мл}
  2. 25\,\text{мл}
  3. 50\,\text{мл}
  4. 150\,\text{мл}
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Given
V_1
120\,\text{мл}
V_2
70\,\text{мл}
Find

\Delta V

  1. \Delta V=V_1-V_2\Delta V=120 - 70\Delta V=50\,\text{мл}

    During compression, the distance between particles changes mainly.

Answer50\,\text{мл}

Position in this set: 15 from 18

Problem #78

Interaction between particles and states of matter

From 100 ml water received ice volume 109 ml. On how much percent grew up volume? consider percentage relatively original volume water.

Answer options

  1. 18\,\text{\%}
  2. 4{,}5\,\text{\%}
  3. 9\,\text{\%}
  4. 27\,\text{\%}
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Given
V_1
100\,\text{мл}
V_2
109\,\text{мл}
Find

r

  1. \Delta V=V_2-V_1\Delta V=109 - 100\Delta V=9\,\text{мл}

    Finding the increment in volume.

  2. r=\frac{\Delta V}{V_1}\cdot100r=\frac{9}{100} \cdot 100r=9\,\text{\%}

    The mass remains the same, but the volume may change.

Answer9\,\text{\%}

Position in this set: 16 from 18

Problem #79

Interaction between particles and states of matter

From 200 ml water received ice volume 218 ml. On how much percent grew up volume? consider percentage relatively original volume water.

Answer options

  1. 18\,\text{\%}
  2. 4{,}5\,\text{\%}
  3. 27\,\text{\%}
  4. 9\,\text{\%}
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Given
V_1
200\,\text{мл}
V_2
218\,\text{мл}
Find

r

  1. \Delta V=V_2-V_1\Delta V=218 - 200\Delta V=18\,\text{мл}

    Finding the increment in volume.

  2. r=\frac{\Delta V}{V_1}\cdot100r=\frac{18}{200} \cdot 100r=9\,\text{\%}

    The mass remains the same, but the volume may change.

Answer9\,\text{\%}

Position in this set: 17 from 18

Problem #80

Interaction between particles and states of matter

B closed vessel mass 100 g are 80 g ice and 120 g water. After melting total ice from vessel poured out 50 g water. Find mass vessel with remaining water.

Answer options

  1. 500\,\text{г}
  2. 125\,\text{г}
  3. 750\,\text{г}
  4. 250\,\text{г}
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Given
m_0
100\,\text{г}
m_1
80\,\text{г}
m_2
120\,\text{г}
m_3
50\,\text{г}
Find

R

  1. m=m_1+m_2m=80 + 120m=200\,\text{г}

    When melting, the mass of the substance is preserved.

  2. M=m-m_3M=200 - 50M=150\,\text{г}

    Subtract the mass of water removed.

  3. R=m_0+MR=100 + 150R=250\,\text{г}

    Add the mass of the empty vessel.

Answer250\,\text{г}

Position in this set: 18 from 18

Problem #81

Interaction between particles and states of matter

B closed vessel mass 150 g are 120 g ice and 180 g water. After melting total ice from vessel poured out 90 g water. Find mass vessel with remaining water.

Answer options

  1. 360\,\text{г}
  2. 720\,\text{г}
  3. 180\,\text{г}
  4. 1080\,\text{г}
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Given
m_0
150\,\text{г}
m_1
120\,\text{г}
m_2
180\,\text{г}
m_3
90\,\text{г}
Find

R

  1. m=m_1+m_2m=120 + 180m=300\,\text{г}

    When melting, the mass of the substance is preserved.

  2. M=m-m_3M=300 - 90M=210\,\text{г}

    Subtract the mass of water removed.

  3. R=m_0+MR=150 + 210R=360\,\text{г}

    Add the mass of the empty vessel.

Answer360\,\text{г}

The theory behind these problems

Work through the topic if a problem will not give in.

Particles and diffusionMolecules, particle motion, Brownian motion and diffusion.Interaction between particles and states of matterAttraction, repulsion and differences between solids, liquids and gases.