All categories

4 years ago

When potassium-40 undergoes decay by electron capture the product is?

1 answer:

andrey2020 [161]4 years ago

4 0

<span>Electron capture decay is a decay process. In this process electron is captured and a proton is converted into a neutron. Hence, mass number remains same but atomic number is lowering by 1 in daughter atom.

K has atomic number 19. Hence daughter atom should have 1</span>9 - 1 = 18<span> as the atomic number. Hence, produced particle is Ar. but the mass number of Ar remains as same as 40.

The equation is,
</span>₁₉⁴⁰K + ₋₁⁰e → ₁₈⁴⁰Ar

You might be interested in

A 10.4 g sample of CaSO4 is found to contain 3.06 g of Ca and 4.89 g of O. Find the mass of sulfur in a sample of CaSO4 with a m

andrey2020 [161]

The mass of sulfur in a sample of CaSO4 with a mass of 65.8 g is 15.50g.

<h3>How to calculate mass of an element in a compound?</h3>

According to this question, a 10.4 g sample of CaSO4 is found to contain 3.06 g of Ca and 4.89 g of O.

This means that the mass of sulfur in the 10.4g of CaSO4 is 10.4g - (3.06g + 4.89g) = 10.4g - 7.95g = 2.45g

Next, we calculate the percent ratio of each element in the compound; CaSO4.

Ca = 3.06g/10.4g × 100 = 29.42%
S = 2.45g/10.4g × 100 = 23.56%
O = 4.89g/10.4g × 100 = 47.02%

According to this question, a sample of CaSO4 with a mass of 65.8 g is given. The mass of each element in this compound is as follows:

Ca = 29.42/100 × 65.8g = 19.36g
S = 23.56/100 × 65.8g = 15.50g
O = 47.02/100 × 65.8g = 30.94g

Therefore, the mass of sulfur in a sample of CaSO4 with a mass of 65.8 g is 15.50g.

Learn more about mass at: brainly.com/question/13672279

#SPJ1

5 0

2 years ago

Explain the reason that maps must change over time

dusya [7]

Explanation:

Because the place on the map can be changed to another shape by natural effects like faulting,folding , volcanism

4 0

3 years ago

Read 2 more answers

Calculate Delta H in KJ for the following reactions using heats of formation:

lozanna [386]

Answer:

$\Delta H\textdegree = -2856.8\;\text{kJ}$ per mole reaction.

$\Delta H\textdegree = -22.3\;\text{kJ}$ per mole reaction.

Explanation:

What is the standard enthalpy of formation $\Delta H_f\textdegree{}$ of a substance? $\Delta H_f\textdegree{}$ the enthalpy change when one mole of the substance is formed from the most stable allotrope of its elements under standard conditions.

Naturally, $\Delta H_f\textdegree{} = 0$ for the most stable allotrope of each element under standard conditions. For example, oxygen $\text{O}_2$ (not ozone $\text{O}_3$ ) is the most stable allotrope of oxygen. Also, under STP $\text{O}_2$ is a gas. Forming $\text{O}_2\;(g)$ from itself does not involve any chemical or physical change. As a result, $\Delta H_f\textdegree{} = 0$ for $\text{O}_2\;(g)$ .

Look up standard enthalpy of formation $\Delta H_f\textdegree{}$ data for the rest of the species. In case one or more values are not available from your school, here are the published ones. Note the state symbols of the compounds (water/steam $\text{H}_2\text{O}$ in particular) and the sign of the enthalpy changes.

$\text{C}_2\text{H}_6\;(g)$ : $-84.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{CO}_2\;(g)$ : $-393.5\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{H}_2\text{O}\;{\bf (g)}$ : $-241.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}\;(s)$ : $-217.9\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}_2\;(s)$ : $-276.6\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{Pb}_3\text{O}_4\;(s)$ : $-734.7\;\text{kJ}\cdot\text{mol}^{-1}$

How to calculate the enthalpy change of a reaction $\Delta H_\text{rxn}$ (or simply $\Delta H$ from enthalpies of formation?

Multiply the enthalpy of formation of each product by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Reactants}))$ to show that this value takes the coefficients into account.
Multiply the enthalpy of formation of each reactant by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Products}))$ to show that this value takes the coefficient into account.
Change = Final - Initial. So is the case with enthalpy changes. $\Delta H_\text{rxn} = \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))$ .

For the first reaction:

$\Sigma (n\cdot \Delta_f(\text{Reactants})) = 4\times (-393.5) + 6\times (-241.8) = -3024.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\Sigma (n\cdot \Delta_f(\text{Products})) = 2\times (-84.0) + 7\times 0 = -168.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\begin{aligned}\Delta H_\text{rxn} &= \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))\\ &= (-3024.8\;\text{kJ}\cdot\text{mol}^{-1}) - (-168.0\;\text{kJ}\cdot\text{mol}^{-1})\\ &= -2856.8\;\text{kJ}\cdot\text{mol}^{-1} \end{aligned}$ .

Try these steps for the second reaction:

$\Delta H_\text{rxn} = -22.3\;\text{kJ}\cdot\text{mol}^{-1}$ .

6 0

3 years ago

What is the empirical formula of a compound containing 90 grams carbon, 11 grams hydrogen, and 35 grams nitrogen? (5 points)

algol13

Answer:

1) C₃H₄N

Explanation:

The empirical formula of a compound is the formula that gives the positive integer ratio of the atoms of the elements in the compound in the simplest form

The mass of carbon in the compound = 90 grams

The molar mass of carbon = 12.011 g/mol

The number of moles of carbon = 90 g/(12.011 g/mol) ≈ 7.4931313 moles

The mass of hydrogen in the compound = 11 grams

The molar mass of hydrogen = 1.00794 g/mol

The number of moles of carbon = 11 g/(1.00794 g/mol) ≈ 10.913348 moles

The mass of nitrogen in the compound = 35 grams

The molar mass of nitrogen = 14.0067 g/mol

The number of moles of carbon = 35 g/(14.0067 g/mol) ≈ 2.49880414 moles

Dividing by the smallest mole ratio gives;

The proportion of carbon, C = 7.4931313/2.49880414 = 2.9987 ≈ 3

The proportion of nitrogen, N = 10.913348 /2.49880414 = 4.367 ≈ 4

The proportion of nitrogen, N = 2.49880414 /2.49880414 = 1

Therefore, the empirical formula of the compound is C₃H₄N.

7 0

3 years ago

If eight atoms of iron react completely with six molecules of oxygen, how many units of iron oxide will be formed?

nevsk [136]

8Fe + 6O₂ -----> 4Fe₂O₃

If eight atoms of iron react completely with six molecules of oxygen, 4 molecules of iron (III) oxide will be formed.

5 0

4 years ago