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4 years ago

You clap your hands and instantly hear an echo. What does this tell you about the size of the room?

Chemistry

2 answers:

STatiana [176]4 years ago

8 0

It is very long and empty. hope this helps.

sveta [45]4 years ago

4 0

A small room...........

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1. A sample of radon has a half-life of 15 hours. What fraction of the sample is left

Hitman42 [59]

I think I learned this

4 0

4 years ago

What do light bulbs emit to produce light?

Tcecarenko [31]

Answer:

light bulbs emit photons to produce light

3 0

3 years ago

Read 2 more answers

A sample of helium gas has a volume of 1.50 L at 159 K and 5.00 atm. When the gas is compressed to 0.200 L at 50.0 atm, the temp

Brums [2.3K]

Answer:

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Explanation:

The given data is as follows.

$V_{1}$ = 1.50 L, $T_{1}$ = 159 K,

$P_{1}$ = 5.00 atm, $V_{2}$ = 0.2 L,

$T_{2}$ = ?, $P_{2}$ = 50.0 atm

And, according to ideal gas equation,

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Hence, putting the given values into the above formula to calculate the value of final temperature as follows.

$\frac{5.0 atm \times 1.50 L}{159 K} = \frac{5 atm \times 0.2 L}{T_{2}}$

$T_{2}$ = 21.27 K

Thus, we can conclude that the final temperature is 21.27 K .

5 0

3 years ago

What are the concentrations of A , A, B , B, and C C at equilibrium if, at the beginning of the reaction, their concentrations a

Elenna [48]

The question is incomplete, here is the complete question:

A reaction

$A+B\rightleftharpoons C$

has a standard free-energy change of -4.88 kJ/mol at 25°C

What are the concentrations of A, B, and C at equilibrium if at the beginning of the reaction their concentrations are 0.30 M, 0.40 M and 0 M respectively?

Answer: The equilibrium concentrations of A, B and C are 0.117 M, 0.217 M, 0.183 M respectively.

Explanation:

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K_c$

where,

$\Delta G^o$ = Standard Gibbs free energy = -4.88 kJ/mol = -4880 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

$K_c$ = equilibrium constant of the reaction

Putting values in above equation, we get:

$-4880J/mol=-(8.3145J/Kmol)\times 298K\times \ln K_c\\\\K_c=7.17$

We are given:

Initial concentration of A = 0.30 M

Initial concentration of B = 0.40 M

Initial concentration of C = 0 M

The chemical reaction follows:

$A+B\rightleftharpoons C$

Initial: 0.30 0.40 0

At eqllm: 0.30-x 0.40-x x

The expression of equilibrium constant for the above reaction follows:

$K_c=\frac{[C]}{[A][B]}$

We are given:

$K_c=7.17$

Putting values in above equation, we get:

$7.17=\frac{x}{(0.30-x)\times (0.40-x)}\\\\x=0.183,0.657$

Neglecting the value of x = 0.657, because change cannot be greater than the initial concentration

So, equilibrium concentration of A = $(0.30-x)=(0.30-0.183)=0.117M$

Equilibrium concentration of B = $(0.40-x)=(0.40-0.183)=0.217M$

Equilibrium concentration of C = $x=0.183M$

Hence, the equilibrium concentrations of A, B and C are 0.117 M, 0.217 M, 0.183 M respectively.

5 0

3 years ago

An ionized helium atom has a mass of 6.6 × 10-27 kg and a speed of 5.3 × 105 m/s. It moves perpendicular to a 0.78-T magnetic fi

arsen [322]

Explanation:

In a magnetic field, the radius of the charged particle is as follows.

r = $\frac{mv}{qB}$

where, m = mass, v = velocity

q = charge, B = magnetic field

Therefore, q will be calculated as follows.

q = $\frac{mv}{rB}$

= $\frac{6.6 \times 10^{-27} \times 5.3 \times 10^{5}}{0.014 m \times 0.78 T}$

= $\frac{34.98 \times 10^{-22}}{0.01092}$

= $3.2 \times 10^{-19} \times \frac{1.0 e}{1.6 \times 10^{-19}C}$

= +2e

Thus, we can conclude that the charge of the ionized atom is +2e.

5 0

3 years ago