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Vedmedyk [2.9K]
3 years ago
8

The speed of sound in air is four times as its speed in steel true or false?​

Chemistry
2 answers:
alexgriva [62]3 years ago
4 0

Answer:

.....false

Explanation:

the particles in the air are packed apart whereas the particles in steel are closely packed and sound travel faster in solids than in gasses..

pychu [463]3 years ago
4 0

Answer:

False

Explanation:

The speed of sound is quickest in mediums that have their molecules close apart.  The speed of sounds is quickest in solids, followed by liquids, and slowest in gasses.  The close the molecules are together, the quicker sound travels.  The sound would travel faster in the steel than in the air.

This is due to the fact that sound waves are compression waves.  This means that they rely on the way molecules collide with one another and act sort of like a spring.  

Fun fact: In the old days, people used to put their ears to the train tracks to see if a train is coming.  Sound moves so fast in metal that they can sense the vibrations miles before the train arrived.

Please mark this as brainliest if this helped you!

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When 500 g of water is cooled from 80.0°C to 10.0°C, how much heat energy is lost?
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I don’t know

Explanation:

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Ghella [55]
I'd say the correct answer is: Noodles rising and falling apart in boiling water.
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Name the 3 seperate metals group
umka21 [38]

Akali Metals

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8 0
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How much heat, in calories, is needed to raise the temperature of 125.0 g of Lead (c lead = 0.130 J/g°C) from 17.5°C to 41.1°C?
pav-90 [236]
95.6 cal
are needed.
Explanation:
Use the following equation:
q
=
m
c
Δ
T
,
where:
q
is heat energy,
m
is mass,
c
is specific heat capacity, and
Δ
T
is the change in temperature.
Δ
T
=
T
final
−
T
initial
Known
m
=
125 g
c
Pb
=
0.130
J
g
⋅
∘
C
T
initial
=
17.5
∘
C
T
final
=
42.1
∘
C
Δ
T
=
42.1
∘
C
−
17.5
∘
C
=
24.6
∘
C
Unknown
q
Solution
Plug the known values into the equation and solve.
q
=
(
125
g
)
×
(
0.130
J
g
⋅
∘
C
)
×
(
24.6
∘
C
)
=
400. J

(rounded to three significant figures)
Convert Joules to calories
1 J
=
0.2389 cal
to four significant figures.
400
.
J
×
0.2389
cal
1
J
=
95.6 cal

(rounded to three significant figures)
95.6 cal
are needed.
8 0
3 years ago
So I saw this question: If 28.0 grams of Pb(NO3)2 react with 18.0 grams of NaI, what mass of PbI2 can be produced? Pb(NO3)2 + Na
Nataly_w [17]

mass of PbI₂ = 27.6606 g

<h3>Further explanation</h3>

Given

Pb(NO₃)₂ + NaI → PbI₂ + NaNO₃

28.0 grams of Pb(NO₃)₂ react with 18.0 grams of NaI

Required

mass of PbI₂

Solution

Balanced equation

Pb(NO₃)₂ + 2NaI → PbI₂ + 2NaNO₃

The principle of a balanced reaction is the number of atoms in the reactants = the number of atoms in the product

mol Pb(NO₃)₂ :

= 28 : 331,2 g/mol

= 0.0845

mol NaI :

= 18 : 149,89 g/mol

= 0.12

Limiting reactant : mol : coefficient

Pb(NO₃)₂ : 0.0845 : 1 = 0.0845

NaI : 0.12 : 2 = 0.06

NaI limiting reactant (smaller ratio)

mol PbI₂ based on NaI

= 1/2 x 0.12 = 0.06

Mass PbI₂ :

= 0.06 x 461,01 g/mol

= 27.6606 g

4 0
3 years ago
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