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3 years ago

3. Write the following isotope in hyphenated form (e.g., "carbon-14”): Kr

Chemistry

2 answers:

irakobra [83]3 years ago

7 0

Answer:

D. Krypton-73

Explanation:

An isotope of an element has the same atomic number and the same number of protons but a different number of neutrons and a different atomic weight. Krypton is the 36th element on the periodic table. It has 36 protons and 48 neutrons. Krypton-73 is one of 33 known isotopes of Krypton and is the only one that actually exists from the list of choices.

Hope that helps.

disa [49]3 years ago

4 0

Answer:

Krypton -73

Explanation:

There are 33 known isotopes of krypton (36Kr) with atomic mass numbers from 69 through 101.

Good luck!

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When backpacking in the wilderness, hikers often boil water to sterilize it for drinking. Suppose that you are planning a backpa

Pavlova-9 [17]

Answer:

2.104 L fuel

Explanation:

Given that:

Volume of water = 35 L = 35 × 10³ mL

initial temperature of water = 25.0 ° C

The amount of heat needed to boil water at this temperature can be calculated by using the formula:

$q_{boiling} = mc \Delta T$

where

specific heat of water c= 4.18 J/g° C

$q_{boiling} = 35 \times 10^{3} \times \dfrac{1.00 \ g}{1 \ mL} \times 4.18 \ J/g^0 C \times (100 - 25)^0 C$

$q_{boiling} = 10.9725 \times 10^6 \ J$

Also; Assume that the fuel has an average formula of C7 H16 and 15% of the heat generated from combustion goes to heat the water;

thus the heat of combustion can be determined via the expression

$q_{combustion} =- \dfrac{q_{boiling}}{0.15}$

$q_{combustion} =- \dfrac{10.9725 \times 10^6 J}{0.15}$

$q_{combustion} = -7.315 \times 10^{7} \ J$

$q_{combustion} = -7.315 \times 10^{4} \ kJ$

For heptane; the equation for its combustion reaction can be written as:

$C_7H_{16} + 11O_{2(g)} -----> 7CO_{2(g)}+ 8H_2O_{(g)}$

The standard enthalpies of the products and the reactants are:

$\Delta H _f \ CO_{2(g)} = -393.5 kJ/mol$

$\Delta H _f \ H_2O_{(g)} = -242 kJ/mol$

$\Delta H _f \ C_7H_{16 }_{(g)} = -224.4 kJ/mol$

$\Delta H _f \ O_{2{(g)}} = 0 kJ/mol$

Therefore; the standard enthalpy for this combustion reaction is:

$\Delta H ^0= \sum n_p\Delta H^0_{f(products)}- \sum n_r\Delta H^0_{f(reactants)}$

$\Delta H^0 =( 7 \ mol ( -393.5 \ kJ/mol) + 8 \ mol (-242 \ kJ/mol) -1 \ mol( -224.4 \ kJ/mol) - 11 \ mol (0 \ kJ/mol))$

$\Delta H^0 = (-2754.5 \ \ kJ - 1936 \ \ kJ+224.4 \ \ kJ+0 \ \ kJ)$

$\Delta H^0 = -4466.1 \ kJ$

This simply implies that the amount of heat released from 1 mol of C7H16 = 4466.1 kJ

However the number of moles of fuel required to burn $7.315 \times 10^{4} \ kJ$ heat released is:

$n_{fuel} = \dfrac{q}{\Delta \ H^0}$

$n_{fuel} = \dfrac{-7.315 \times 10^{4} \ kJ}{-4466.1 \ kJ}$

$n_{fuel} = 16.38 \ mol \ of \ C_7 H_{16$

Since number of moles = mass/molar mass

The mass of the fuel is:

$m_{fuel } = 16.38 mol \times 100.198 \ g/mol}$

$m_{fuel } = 1.641 \times 10^{3} \ g$

Given that the density of the fuel is = 0.78 g/mL

and we know that :

density = mass/volume

therefore making volume the subject of the formula in order to determine the volume of the fuel ; we have

volume of the fuel = mass of the fuel / density of the fuel

volume of the fuel = $\dfrac{1.641 \times 10^3 \ g }{0.78 g/mL} \times \dfrac{L}{10^3 \ mL}$

volume of the fuel = 2.104 L fuel

3 0

3 years ago

La Pokebola del gráfico realiza un MCU con una rapidez angular de 2π 9 rad s ⁄ . Determine el tiempo que emplea para ir desde A

ddd [48]

Fuvghbgbhj kkiggggbb ghhhb

7 0

2 years ago

In the laboratory you dissolve 19.4 g of potassium acetate in a volumetric flask and add water to a total volume of 125 mL. What

aleksandrvk [35]

The molarity of the potassium acetate solution given the data is 1.584 M

<h3>What is molarity? </h3>

This is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:

Molarity = mole / Volume

<h3>How to determine the mole of CH₃COOK</h3>

Mass of CH₃COOK = 19.4 g
Molar mass of CH₃COOK = 98 g/mol
Mole of CH₃COOK =?

Mole = mass / molar mass

Mole of CH₃COOK = 19.4 / 98

Mole of CH₃COOK = 0.198 mole

<h3>How to determine the molarity of CH₃COOK</h3>

Mole of CH₃COOK = 0.198 mole
Volume = 125 mL = 125 / 1000 = 0.125 L
Molarity of CH₃COOK = ?

Molarity = mole / Volume

Molarity of CH₃COOK = 0.198 / 0.125

Molarity of CH₃COOK = 1.584 M

Learn more about molarity:

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5 0

2 years ago

White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white

Alex_Xolod [135]

The pH = 2.41

<h3>Further explanation</h3>

Given

5.0% by mass solution of acetic acid

the density of white vinegar is 1.007 g/cm3

Required

Solution

Molarity of solution :

$\tt M=\dfrac{\%mass\times \rho\times 10}{MW~acetic~acid}\\\\M=\dfrac{5\times 1.007\times 10}{60}\\\\M=0.839$

Ka for acetic acid = 1.8 x 10⁻⁵

[H⁺] for weak acid :

$\tt [H^+]=\sqrt{Ka.M}$

Input the value :

$\tt [H^+]=\sqrt{1.8\times 10^{-5}\times 0.839}\\\\(H^+]=0.00388=3.88\times 10^{-3}\\\\pH=3-log~3.88=2.41$

7 0

3 years ago

What are the charges of families 1, 2, & 13 (boron family) if/when they become Ions

Natalka [10]

Answer:

Explanation:

Group one elements are alkali metals. All alkali metal have one valance electron. They loses their one valance electron and from cation with charge of +1.

Charges on group one.

Hydrogen = +1

Lithium = +1

Sodium = +1

Potassium = +1

Rubidium = +1

Cesium = +1

Francium = +1

Group two elements are alkaline earth metals. All alkaline earth metal have two valance electron. They loses their two valance electron and from cation with charge of +2.

Charges on group two.

Beryllium = +2

Magnesium = +2

Calcium = +2

Strontium = +2

Barium= +2

Radium = +2

Group 13 elements are boron family. All elements have three valance electrons. They loses their three valance electron and from cation with charge of +3.

Charges on group 13.

Boron = +3

Aluminium = +3

Gallium = +3

Indium = +3

Thallium= +3

Group 13 elements are also shows +1 charge by losing one valance electron.

8 0

3 years ago