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krek1111 [17]
4 years ago
9

Which notation is used to represent a beta particle?

Chemistry
2 answers:
Lapatulllka [165]4 years ago
7 0
The beta particle is represented by the Greek letter <span>β. The beta particle is one of the particles that are emitted during radioactive decay. Beta particles are emitted in the process of beta decay. A beta particle can be in form of an electron or a positron. </span><span />
Fofino [41]4 years ago
6 0

the answer is d for plato

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A sample of neon occupies a volume of 461 mL at 1 atm and 273 K. What will be the volume of the neon when the pressure is reduce
Softa [21]

Answer:

0.501 L

Explanation:

To solve this problem we will use Boyle,s Law. According to this law "The volume of given amount of gas is inversely proportional to applied pressure at constant temperature".

                           V∝ 1/P

                           V= K/P

                            VP=K

Here the K is proportionality constant.

so,

             P1V1 = P2V2

P= pressure

V= volume

Given data:

P1= 1 atm

V1= 461 mL

P2= 0.92 atm

V2= ? (L)

To solve this problem we have to convert the mL into L first.

1 L = 1000 mL

461/1000= 0.461 L

Now we will put the values in the equation,

P1V1 = P2V2

V2= P1V1/ P2

V2= 1 atm × 0.461 L / 0.92 atm

V2= 0.501 L

4 0
4 years ago
011 10.0 points
Sphinxa [80]

Answer:

-58.3^{\circ}C

Explanation:

The frequency and the wavelength of a wave are related by the equation

v=f\lambda

where

v is the speed

f is the frequency

\lambda is the wavelength

For the sound wave in this problem, we have

f = 634 Hz (frequency)

\lambda=0.47 m (wavelength)

Therefore, the speed of the wave is

v=(634)(0.47)=298 m/s

The speed of sound in air depends on the temperature according to the equation

v=333+0.6 T

where T is the temperature.

In this problem, we know the speed, so we can calculate the temperature:

T=\frac{v-333}{0.6}=\frac{298-333}{0.6}=-58.3^{\circ}C

7 0
3 years ago
Being able to roll the tongue is dominant over not being able to roll the tongue.
-Dominant- [34]

Answer:

Its 0

Explanation:

7 0
3 years ago
Read 2 more answers
An acidified solution was electrolyzed using copper electrodes. A constant current of 1.18 A caused the anode to lose 0.584 g af
Alexxx [7]

Answer:

\boxed{\text{(a) 209 mL; (b) } 6.09 \times 10^{23}}

Explanation:

(a) Gas produced at cathode.

(i). Identity

The only species known to be present are Cu, H⁺, and H₂O.

Only the H⁺ and H₂O can be reduced.

The corresponding reduction half reactions are:

(1) 2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻;     E° = -0.8277 V

(2) 2H⁺ +2e⁻ ⇌ H₂;                     E° =  0.0000 V

Two important points to remember when using a table of standard reduction potentials:

  • The higher up a species is on the right-hand side, the more readily it will lose electrons (be oxidized).
  • The lower down a species is on the left-hand side, the more readily it will accept electrons (be reduced}.

H⁺ is below H₂O, so H⁺ is reduced to H₂.

The cathode reaction is 2H⁺ +2e⁻ ⇌ H₂, and the gas produced at the cathode is hydrogen.

(ii) Volume

a. Anode reaction

The only species that can be oxidized are Cu and H₂O.

The corresponding half reactions  are:

(3) Cu²⁺ + 2e⁻ ⇌ Cu;                E° =  0.3419 V

(4) O₂ + 4H⁺ + 4e⁻ ⇌ 2H₂O     E° =   1.229   V

Cu is above H₂O, so Cu is more easily oxidized.

The anode reaction is Cu ⇌ Cu²⁺ + 2e⁻.

b. Overall reaction:

Cu           ⇌ Cu²⁺ + 2e⁻

<u>2H⁺ +2e⁻ ⇌ H₂            </u>        

Cu + 2H⁺ ⇌ Cu²⁺ + H₂

c. Moles of Cu lost

n_{\text{Cu}} = \text{0.584 g } \times \dfrac{\text{1 mol}}{\text{63.55 g}} = 9.190 \times 10^{-3}\text{ mol Cu}

d. Moles of H₂ formed

n_{\text{H}_{2}}} = 9.190 \times 10^{-3}\text{ mol Cu} \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol Cu}} =9.190 \times 10^{-3}\text{ mol H}_{2}

e. Volume of H₂ formed

Volume of 1 mol at STP (0 °C and  1 bar) = 22.71 mL

V = 9.190 \times 10^{-3}\text{ mol}\times \dfrac{\text{22.71 L}}{\text{1 mol}}  = \text{0.209 L} = \boxed{\textbf{209 mL}}

(b) Avogadro's number

(i) Moles of electrons transferred

\text{Moles of electrons} = 9.190 \times 10^{-3}\text{ mol Cu}\times \dfrac{\text{2 mol electrons}}{\text{1 mol Cu}}\\\\\\= \text{0.018 38 mol electrons}

(ii) Number of coulombs

Q  = It  

Q = \text{1.18 C/s} \times 1.52 \times 10^{3} \text{ s} = 1794 C

(iii). Number of electrons

n = \text{ 1794 C} \times \dfrac{\text{1 electron}}{1.6022 \times 10^{-19} \text{ C}} = 1.119 \times 10^{22} \text{ electrons}

(iv) Avogadro's number

N_{\text{A}} = \dfrac{1.119 \times 10^{22} \text{ electrons}}{\text{0.018 38 mol}} = \boxed{6.09 \times 10^{23} \textbf{ electrons/mol}}

6 0
3 years ago
How is carbon moved from the hydrosphere to the atmosphere?
inessss [21]

Answer:

Explained

Explanation:

the various ways of input of CO_2 into the atmosphere.

1.Dissolved  CO 2 in the ocean is released back into the atmosphere by heating ocean surface water

2.Plant and animal respiration, an exothermic reaction involving the breakdown into CO 2 and water of organic molecules.

3. Degradation of fungi and bacteria which are responsible for breaking down carbon compounds in dead animals and plants(fossils) and convert carbon into CO2 when oxygen or methane is present.

4.Combustion of organic matter (that includes deforestation and combustion of fossil fuels) oxidizing to produce CO2;

5. Cement production when calcium carbonate (limestone) is heated to produce calcium oxide(lime), cement component, and CO2 are released;

4 0
3 years ago
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