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3 years ago

Alpha and beta particles and gamma rays are examples of:

Chemistry

2 answers:

Ann [662]3 years ago

7 0

C. Alpha and beta particles, along with gamma rays are examples of ionizing radiation.

Semmy [17]3 years ago

5 0

Answer : The correct option is, (c) ionizing radiation

Explanation :

Ionizing radiation : it is defined as the energy that is released by the atoms which travels in the form of electromagnetic waves or particles.

The electromagnetic waves are gamma-ray, X-rays, U.V rays, infrared ray, microwave and radio-wave.

The electromagnetic particles are alpha, beta and neutrons.

Isotope : It is defined as the element that have the same number of protons but have the different number of neutrons of each of the atom.

Hence, the alpha and beta particles and gamma rays are examples of ionizing radiation.

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Which of the solutions below is a weak electrolyte? nacl(aq) koh(aq) hcl(aq) h2so3(aq) all of these?

Deffense [45]

Answer:-

H2SO3 (aq)

Explanation:-

NaCl, KOH and HCl ionic compounds.

Ionic compounds dissociate into ions in water.

Ions can carry electricity in water. So NaCl, KOH and HCl are all strong electrolytes.

H2SO3 is not a strong acid and dissociates to a small extent into ions in water.

Hence number of ions formed is less. Hence it is a weak electrolyte.

5 0

3 years ago

A single atom of an element has 21 neutrons, 20 electrons, and 20 protons. Which element is it?

jeyben [28]

Answer:

Calcium

Explanation:

Clacium has atomic number 20. The atomic number. This means that it has 20 protons and 20 electrons. it has 21 neutrons.

5 0

3 years ago

Be sure to answer all parts. Write the balanced equations corresponding to the following rate expressions: a) rate = − 1 3 Δ[CH4

Alinara [238K]

Answer : The balanced equations will be:

(a) $3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH$

(b) $2N_2O_5\rightarrow 2N_2+5O_2$

(c) $2H_2+2CO_2+O_2\rightarrow 2H_2CO_3$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

Now we have to determine the balanced equations corresponding to the following rate expressions.

(a) $Rate=-\frac{1}{3}\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[H_2O]}{dt}=-\frac{d[CO_2]}{dt}=+\frac{1}{4}\frac{d[CH_3OH]}{dt}$

The balanced equations will be:

$3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH$

(b) $Rate=-\frac{1}{2}\frac{d[N_2O_5]}{dt}=+\frac{1}{2}\frac{d[N_2]}{dt}=+\frac{1}{5}\frac{d[O_2]}{dt}$

The balanced equations will be:

$2N_2O_5\rightarrow 2N_2+5O_2$

(c) $Rate=-\frac{1}{2}\frac{d[H_2]}{dt}=-\frac{1}{2}\frac{d[CO_2]}{dt}=-\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[H_2CO_3]}{dt}$

The balanced equations will be:

$2H_2+2CO_2+O_2\rightarrow 2H_2CO_3$

4 0

3 years ago

What type of mixture is throughly mixing powdered lemonade with water

vagabundo [1.1K]

Answer:

I believe it would be a solution I could be wrong tho

Explanation:

8 0

2 years ago

Read 2 more answers

Titanium has five common isotopes: 46Ti (8.0%), 47Ti (7.8%), 48Ti (73.4%), 49Ti (5.5%), 50Ti (5.3%). What is the average atomic

NNADVOKAT [17]

Hey there!:

Isotopes : abundance :

46 Ti 8.0%

47 Ti 7.8 %

48 Ti 73.4 %

49 Ti 5.5 %

50 Ti 5.3 %

Weighted average = ∑ Wa * % / 100

Therefore:

( 46 * 8.0) + (47 * 7.8 ) + (48 * 73.4 ) + ( 49 * 5.5 ) + ( 50*5.3 ) / 100 =

4792.3 / 100

= 47.923 a.m.u

Hope that helps!

7 0

3 years ago