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2 years ago

The nucleotides I-131 and I-133 are classified as?

Chemistry

2 answers:

Arada [10]2 years ago

6 0

Answer : Isotopes.

Explanation : The nuclides of I-131 and I-133 are classified as isotopes of iodine.

An isotope is each of two or more forms of the same element that contains equal numbers of protons but differ in numbers of neutrons in their atomic nuclei, which makes them differ in relative atomic mass.

This makes them radioactive too.

erastovalidia [21]2 years ago

3 0

Answer: These two nucleotides are termed as isotopes.

Explanation: Isotopes are the species of the same element having same number of protons and electrons but different number of neutrons.

Atomic number remains same but their atomic masses are different.

These two nucleotides are the isotopes of Iodine having atomic number 53.

$^{131}_{53}I$ and $^{133}_{53}I$ are classified as Isotopes of Iodine.

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In which 1.0 gram sample are particles arranged in a rystal structure?

lilavasa [31]

The Options are as follow,

(1) CaCl₂ (s) (3) CH₃OH (l)

(2) C₂H₆ (g) (4) Cal₂ (aq)

Answer:

Option-1 is the correct answer.

Explanation:

As we know crystal formation is the property of solids. Therefore, in given options we are given with four different states of matter.

Option A, CaCl₂ is in a solid state , so it can exist in crystal form.

Option 2, C₂H₆ (Ethane) is in gas form, so it cannot form crystals.

Option 3, CH₃OH (Methanol) is present in liquid form, so it fails to form crystals.

Option 4, CaI₂, it is dissolved in water, Hence, it is in aqueous state, Therefore it also lacks crystal structure.

5 0

2 years ago

A penny has a mass of

Ahat [919]

You did not include the questions.

I did some research and found the questions:

 What is the mass of 1 mole of pennies? How many moles of pennies have a mass equal to the mass of the moon?

Solutions:

1) mass of 1 mole of pennies

Data: mass of 1 penny = 2.50 g

1 mole = 6.022 * 10^ 23 units

Proportion:

1 penny 6.022 * 10^23 penny
-------------- = ----------------------------
 2.50 g x

Solve: x = 6.022 * 10^23 penny * 2.50g / 1 penny = 15.055* 10^23

Since 2.50 has 3 significant figures, the answer must use 3 significant figures => x = 15.1 * 10^ 23 g = 1.51 * 10^24 g

Answer: 1 mol of pennies have a mass of 1.51 * 10^24 g

2) How many moles of pennies have a mass equal to the same mass of the Moon

Convert the mass of the Moon grams: 7.35 * 10^22 kg = 7.35 * 10^ 25 g

 1 mol x
---------------------- = ----------------------
1.51 * 10^ 24g 7.35 * 10^ 25 g

=> x = 7.35 * 10^ 25 g * 1 mol / (1.51 * 10^24 g)= 48.7 mol

Answer: 48.7 mol

5 0

2 years ago

Read 2 more answers

For the reaction 2N2O5(g) <---> 4NO2(g) + O2(g), the following data were colected:

KonstantinChe [14]

Answer:

a) The reaction is first order, that is, order 1. Option C is correct.

b) The half life of the reaction is 23 minutes. Option B is correct

c) The initial rate of production of NO2 for this reaction is approximately = (3.7 × 10⁻⁴) M/min. Option has been cut off.

Explanation:

First of, we try to obtain the order of the reaction from the data provided.

t (minutes) [N2O5] (mol/L)

0 1.24x10-2

10 0.92x10-2

20 0.68x10-2

30 0.50x10-2

40 0.37x10-2

50 0.28x10-2

70 0.15x10-2

Using a trial and error mode, we try to obtain the order of the reaction. But let's define some terms.

C₀ = Initial concentration of the reactant

C = concentration of the reactant at any time.

k = rate constant

t = time since the reaction started

T(1/2) = half life

We Start from the first guess of zero order.

For a zero order reaction, the general equation is

C₀ - C = kt

k = (C₀ - C)/t

If the reaction is indeed a zero order reaction, the value of k we will obtain will be the same all through the set of data provided.

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = (0.0124 - 0.0092)/10 = 0.00032 M/min

At t = 20 minutes, C = 0.0068 M

k = (0.0124 - 0.0068)/20 = 0.00028 M/min

At t = 30 minutes, C = 0.0050 M

k = (0.0124 - 0.005)/30 = 0.00024 M/min

It's evident the value of k isn't the same for the first 3 trials, hence, the reaction isn't a zero order reaction.

We try first order next, for first order reaction

In (C₀/C) = kt

k = [In (C₀/C)]/t

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = [In (0.0124/0.0092)]/10 = 0.0298 /min

At t = 20 minutes, C = 0.0068 M

k = 0.030 /min

At t = 30 minutes, C = 0.0050 M

k = 0.0303

At t = 40 minutes

k = 0.0302 /min

At t = 50 minutes,

k = 0.0298 /min

At t = 60 minutes,

k = 0.031 /min

This shows that the reaction is indeed first order because all the answers obtained hover around the same value.

The rate constant to be taken will be the average of them all.

Average k = 0.0302 /min.

b) The half life of a first order reaction is related to the rate constant through this relation

T(1/2) = (In 2)/k

T(1/2) = (In 2)/0.0302

T(1/2) = 22.95 minutes = 23 minutes.

c) The initial rate of production of the product at the start of the reaction

Rate = kC (first order)

At the start of the reaction C = C₀ = 0.0124M and k = 0.0302 /min

Rate = 0.0302 × 0.0124 = 0.000374 M/min = (3.74 × 10⁻⁴) M/min

3 0

2 years ago

2. Suggest four ways in which the concentration of PH3 could be increased in an equilibrium described by the following equation:

Nesterboy [21]

Answer

increase in temperature

decrease in pressure

continuous removal of PH3
adding more of P into the system

Explanation:

In the reaction P4(g)+6H2(g) ⇌ 4PH3(g);

The effect of temperature on equilibrium has to do with the heat of reaction. Recall that for an endothermic reaction, heat is absorbed in the reaction, and the value of ΔH is positive. Thus, for an endothermic reaction, we can picture heat as being a reactant:

heat+A⇌BΔH=+

Since the reaction is endothermic reaction, heat is a absorbed. Decreasing the temperature will shift the equilibrium to the left, while increasing the temperature will shift the equilibrium to the right forming more of PH3.

According to Le Chatelier’s principle, adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products. In the same Way, reducing the concentration of the product will also shift equilibrium to the right continually forming PH3 as it is removed.

4 0

2 years ago

Identify the MOs that react to form cyclohexene. HOMO of 1,3-butadiene and LUMO of ethylene LUMO of 1,3-butadiene and LUMO of et

BARSIC [14]

Answer:

HOMO of 1,3-butadiene and LUMO of ethylene

HOMO of ethylene LUMO of 1,3-butadiene

Explanation:

1,3 - butadiene underogoes cycloaddition reaction with ethylene to give cyclohexene.

According to Frontier molecular orbital theory HOMO of 1,3 butadiene and LUMO of ethylene and HOMO of ethylene and LUMO of ethylene underoges (4 + 2) in thermal or photochemical condition.

6 0

2 years ago