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3 years ago

What types of radiation cause the parent isotope to change into a different element?

Chemistry

1 answer:

kvv77 [185]3 years ago

3 0

Answer:

Beta decay is most common in elements with a high neutron to proton ratio. Gamma decay follows the form: In gamma emission, neither the atomic number or the mass number is changed. A high energy gamma ray is given off when the parent isotope falls into a lower energy state.

Explanation:

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You might be interested in

How is an endothermic reaction identified?

Naily [24]

Answer:

Endothermic is identified if the process requires heat as a reactant, since endothermic means "intake of heat", so it has to take in some type of heat.

7 0

3 years ago

A student prepares a solution of sodium chloride by dissolving 116.9 g of NaCl into enough water to make 1.00 L of solution. How

sertanlavr [38]

Answer: The molarity of the solution is 2.00 M

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L = 1.00 L

moles of solute = $\frac{\text {given mass}}{\text {molar mass}}=\frac{116.9g}{58.5g/mol}=2.00moles$

$Molarity=\frac{2.00mol}{1.00L}=2.00M$

Thus the molarity of the solution is 2.00 M

8 0

4 years ago

The rate law for the reaction 2A + B →C is –rA= kACA2CBwithkA= 25 (L/mol)2sec. What arekBandkC?

givi [52]

Explanation:

The given reaction equation is as follows.

$2A + B \rightarrow C$

So, rate constants for different reactants and products written as follows.

$\frac{-k_{A}}{\text{stoichiometric coefficient of A}} = \frac{-k_{B}}{\text{stoichiometric coefficient of B}} = \frac{k_{C}}{\text{stoichiometric coefficient of C}}$

As per the reaction equation, the stoichiometric coefficients of reactants and products are as follows.

A = -2

B = -1

C = 1

Therefore,

$\frac{-k_{A}}{\text{stoichiometric coefficient of A}} = \frac{-k_{B}}{\text{stoichiometric coefficient of B}} = \frac{k_{C}}{\text{stoichiometric coefficient of C}}$

$\frac{-k_{A}}{-2} = \frac{-k_{B}}{-1} = \frac{k_{C}}{1}$

$\frac{-k_{A}}{-2} = k_{B} = k_{C}$

Hence, $k_{B} = k_{C}$ = $\frac{25}{2} (L/mol)^{2}$

= 12.5 $(L/mol)^{2}$

Thus, we can conclude that $k_{B}$ and $k_{C}$ are 12.5 $(L/mol)^{2}$ .

5 0

3 years ago

Be sure to answer all parts.

djyliett [7]

The number of moles contained in each of the question are a) 3.935 and b) 0.001165 mol.

<h3>What are moles?</h3>

Moles are a quantity or amount of a substance. Denoted by n.

To calculate moles, mass is divided by the molar mass

n = m/M

mass is divided by the molar mass

a) 230 grams of NaCl

n = m/M

molar mass of NaCl is 58.44 g/mol

n = 230 / 58.44 = 3.935 mol

b) 0.210 g of aspirin

n = m/M

molar mass of aspirin is 180.158 g/mol

n = 0.210 / 180.158 = 0.001165 mol

Thus, the moles are a) 3.935 and b) 0.001165 mol.

Learn more about moles

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4 0

2 years ago

2AgNO3 + BaCl2 → 2AgCl + Ba(NO3)2

miv72 [106K]

<h3>Answer:</h3>

4 g AgCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN] 2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂

[Given] 5.0 g AgNO₃

<u>Step 2: Identify Conversions</u>

[Reaction - Stoich] 2AgNO₃ → 2AgCl

Molar Mass of Ag - 107.87 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol

Molar Mass of AgCl - 107.87 + 35.45 = 143.32 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 5.0 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{2 \ mol \ AgCl}{2 \ mol \ AgNO_3})(\frac{143.22 \ g \ AgCl}{1 \ mol \ AgCl})$
Multiply/Divide: $\displaystyle 4.21533 \ g \ AgCl$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

4.21533 g AgCl ≈ 4 g AgCl

3 0

3 years ago