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3 years ago

9. Write a balanced nuclear equation for the following: The isotope Strontium-90 decays by Q-

Chemistry

1 answer:

Sergio039 [100]3 years ago

4 0

Answer: $_{38}^{90}\textrm{Sr}\rightarrow _{36}^{86}\textrm{Kr}+_2^4\textrm{He}$

Explanation:

Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.

General representation of an element is given as: $_Z^A\textrm{X}$

where,

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

General representation of alpha decay :

$_Z^A\textrm{Sr}\rightarrow _{Z-2}^{A-4}Y+_2^4\alpha$

The balanced nuclear equation when the isotope Strontium-90 decays by Q- decay is :

$_{38}^{90}\textrm{Sr}\rightarrow _{36}^{86}\textrm{Kr}+_2^4\textrm{He}$

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Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacc

Ira Lisetskai [31]

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is $C_5H_7N$

(b) The empirical formula for the given compound is $C_4H_5N_2O$

Explanation:

Part A: nicotine

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.23 moles.

For Carbon = $\frac{6.17}{1.23}=5.01\approx 5$

For Hydrogen = $\frac{8.70}{1.23}=7.07\approx 7$

For Nitrogen = $\frac{1.23}{1.23}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

The empirical formula for the given compound is $C_5H_7N_1=C_5H_7N$

Part B: caffeine

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.03 moles.

For Carbon = $\frac{4.12}{1.03}=4$

For Hydrogen = $\frac{5.19}{1.03}=5.03\approx 5$

For Nitrogen = $\frac{2.06}{1.03}=2$

For Nitrogen = $\frac{1.03}{1.03}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is $C_4H_5N_2O_1=C_4H_5N_2O$

6 0

3 years ago

How do I find the number of moles and molar mass of Gas A and Gas B with the information provided?

drek231 [11]

Answer:

Explanation:

hope this helped sorry if its wrong!

3 0

2 years ago

Consider this chemical reaction. What chemical is reduced? CH3OH + NAD --> CH2O + NADH

nikklg [1K]

Answer:

CH3OH and NADH

Explanation:

The given chemical reaction is an redox reaction in which reduction and oxidation take place.

In the process of oxidation: electrons are loss while in the process of reduction: electrons are gained.

In the given redox reaction: CH3OH + NAD --> CH2O + NADH

NAD is reduced to NADH as NADH gains one hydrogen electron while CH3OH (methanol) is oxidized to CH2O (methanal) by losing electrons.

So, CH3OH (methanol) and NADH are the reduced forms while NAD and CH2O (methanal) are oxidized forms.

5 0

3 years ago

What is the coefficient for hydrogen in the balanced equation for the reaction of solid molybdenum(iv) oxide with gaseous hydrog

anastassius [24]

The coefficient for hydrogen in the balanced equation of solid molybdenum(iV) oxide with gaseous hydrogen is 2

Explanation

Coefficient is defined to as a number in front of a chemical formula in a balanced chemical equation.

The reaction of molybdenum (iv) oxide with gaseous hydrogen is as below,

MoO2 + 2 H2→ Mo +2 H2O

From balanced equation above the coefficient for H2 is 2 since the number in front of H2 is 2

3 0

3 years ago

A chemist wants to make 6.5 L of a .350M CaCl2 solution. What mass of CaCl2(in g) should the chemist use?

kotykmax [81]

First M stands for Molarity which is (moles of solute) / (Liters of solution). we also know that moles = (mass) / (molar mass). so we can form some equations here. We know:
Molarity (M) = moles (mol) / Liters (L)
moles (mol) = (mass) / (molar mass)

we can substitute the (mass) / (molar mass) for (moles) and get:
M = [(mass) / (molar mass)] / Liters

we can now isolate mass and get
M * Liters * molar mass = mass

now we need to find the molar mass of CaCl2 which is 110.98 g/mol

plug the values in and get
.350M * 6.5L * 110.98 g/mol = mass

mass = 252.4795g however the 6.5L has only 2 sig figs so i would say

mass CaCl2 = 2.5 * 10 ^2 g

5 0

3 years ago

Read 2 more answers