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3 years ago

Which is an image of a scintillation counter?

Chemistry

2 answers:

Natali [406]3 years ago

5 0

Answer:

The answer is D.

Explanation:

I just took the test.

dimulka [17.4K]3 years ago

3 0

D is a scintillation counter.

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I want to know which ones are molecular equation, complete ionic equation and net ionic equation

NNADVOKAT [17]

Answer:

The molecular equations are:

1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)

2. Ba(NO₃)₂ (aq) + K₂SO₄ (aq) + BaSO₄ (s) + 2 KNO₃ (aq)

The complete ionic equations are:

1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)

2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)

The net ionic equations are:

1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)

2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)

Explanation:

A molecular equation is a balanced chemical equation which shows the reacting species as molecules rather than as componenet ions in their compounds with subscripts written beside the molecules to indicate the state in which they occur in the chemical reaction.

An ionic equation expresses the reacting species as components ions in a chemical reation. All the ions and molecules reacting are shown.

In a net ionic equation, the ions which remain in the ionic state also known as spectator ions are not written as part of the equation.

From the given attachment;

The molecular equations are:

1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)

2. Ba(NO₃)₂ (aq) + K₂SO₄ (aq) + BaSO₄ (s) + 2 KNO₃ (aq)

The complete ionic equations are:

1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)

2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)

The net ionic equations are:

1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)

2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)

8 0

3 years ago

Select the correct answer.

Alenkasestr [34]

Answer:

Explanation:

The unit used to measure atomic mass is the atomic mass unit (amu). A single amu is equivalent to 1/12 the mass of an atom from the carbon-12 isotopIsotopes with different numbers of protons and neutrons will have an actual mass slightly different from the atomic mass calculated in atomic mass units.

4 0

2 years ago

Read 2 more answers

What type of energy transformation takes place during cellular respiration?

kirill [66]

Answer:

chemical energy transgormation takes place during cellular respiration

8 0

3 years ago

QUE VENTAJAS TIENE PARA LA PLANTA Y EL CULTIVO LA PERMANENCIA DEL POTASIO COMO CATION MONOVALENTE (K+)?

yarga [219]

Explanation:

Son muchos los beneficios del catión monovalente (K+) en las plantas y cultivos, mencionaremos algunos:

-Interviene en el proceso de apertura y cierre de los estomas; el catión (k+) ayuda a que los estomas puedan realizar correctamente el proceso de la fotosíntesis es decir la absorción del dióxido de carbono para producir oxígeno y agua; este proceso ayuda a que la planta pueda fabricar su propio alimento es decir transformar materiales inorgánicos en orgánicos como azúcares y carbohidratos que son aprovechables por la planta para realizar diversas funciones catalíticas como la respiración celular,la fosforilación y la síntesis de proteínas; este proceso también contribuye a que la planta pueda almacenar y retener suficiente agua lo cual ayuda a evitar el estrés hídrico en la planta.

Por lo tanto si la planta tiene una buena concentración de potasio como catión monovalente(k+); es posible que tenga un buen crecimiento y desarrollo radicular ,así mismo la calidad de los cultivos será muy buena,seran cultivos muy nutritivos y saludables. Algunos cultivos que necesitan una alta cantidad de potasio son los cítricos y el banano.

6 0

3 years ago

Ethylene (CH2CH2) is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about 8.0 x 10

mylen [45]

Answer: The percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

Explanation:

We are given:

Initial partial pressure or ethane = 24.0 atm

The chemical equation for the dehydration of ethane follows:

$C_2H_6(g)\rightleftharpoons C_2H_4(g)+H_2(g)$

Initial: 24.0

At eqllm: 24-x x x

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{C_2H_4}\times p_{H_2}}{p_{C_2H_6}}$

We are given:

$K_p=0.040$

Putting values in above expression, we get:

$0.040=\frac{x\times x}{24-x}\\\\x^2+0.04x-0.96=0\\\\x=0.96,-1$

Neglecting the value of x = -1 because partial pressure cannot be negative.

So, partial pressure of hydrogen gas at equilibrium = x = 0.96 atm

Partial pressure of ethylene gas at equilibrium = x = 0.96 atm

Partial pressure of ethane gas at equilibrium = (24-x) = (24 - 0.96) atm = 23.04 atm

To calculate the number of moles, we use the equation given by ideal gas, which follows:

$PV=nRT$ .........(1)

To calculate the mass of a substance, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ..........(2)

For ethane gas:

We are given:

$P=23.04atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$23.04atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{23.04\times 30.0}{0.0821\times 1073}=7.85mol$

We know that:

Molar mass of ethane gas = 30 g/mol

Putting values in equation 2, we get:

$7.85mol=\frac{\text{Mass of ethane gas}}{30g/mol}\\\\\text{Mass of ethane gas}=(7.85mol\times 30g/mol)=235.5g$

For ethylene gas:

We are given:

$P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol$

We know that:

Molar mass of ethylene gas = 28 g/mol

Putting values in equation 2, we get:

$0.33mol=\frac{\text{Mass of ethylene gas}}{28g/mol}\\\\\text{Mass of ethylene gas}=(0.33mol\times 28g/mol)=9.24g$

For hydrogen gas:

We are given:

$P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol$

We know that:

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 2, we get:

$0.33mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(0.33mol\times 2g/mol)=0.66g$

To calculate the mass percentage of ethylene in equilibrium gas mixture, we use the equation:

$\text{Mass percent of ethylene gas}=\frac{\text{Mass of ethylene gas}}{\text{Mass of equilibrium gas mixture}}\times 100$

Mass of equilibrium gas mixture = [235.5 + 9.24 + 0.66] = 245.4 g

Mass of ethylene gas = 9.24 g

Putting values in above equation, we get:

$\text{Mass percent of ethylene gas}=\frac{9.24g}{245.5g}\times 100=3.76\%$

Hence, the percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

5 0

3 years ago

Which is an image of a scintillation counter?﻿

Which is an image of a scintillation counter?