Calculate the average atomic mass for X

Calculate The Average Atomic Mass For X

Answers

Answer 1

Answer:

39.0229 amu

Explanation:

Hello there!

In this case, according to given information, the idea here is to multiply the percent abundance by the mass number of each isotope and then add them all together as shown below:

[tex]=0.0967*38+0.7868*39+0.1134*40+0.0031*41\\\\=3.6746+30.6852+4.536+0.1271\\\\=39.0229amu[/tex]

Regards!


Related Questions

Which procedure could a student use to examine an intensive property of a rectangular block of wood
Find the mass.
Record the length. Measure the volume. Determine the density.

Answers

Answer:

density

Explanation:

The procedure that the student could use to examine an intensive property of a rectangular block of wood is to determine its density. Density is intensive because it is the ration between the mass and the volume.

Answer: find the mass option A

Explanation:

Que es la actividad física y en qué mejora

Answers

La actividad física regular puede mejorar su fuerza muscular y aumentar su resistencia. El ejercicio proporciona oxígeno y nutrientes a sus tejidos y ayuda a que su sistema cardiovascular funcione de manera más eficiente. Y cuando la salud de su corazón y pulmones mejoran, tiene más energía para hacer frente a las tareas diarias. Encantado de ayudarle

If 0.250 L of a 5.90 M HNO₃ solution is diluted to 2.00 L, what is the molarity of the new solution?

Answers

Answer:

0.74 M

Explanation:

From the question given above, the following data were obtained:

Molarity of stock solution (M₁) = 5.90 M

Volume of stock solution (V₁) = 0.250 L

Volume of diluted solution (V₂) = 2 L

Molarity of diluted solution (M₂) =?

The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:

M₁V₁ = M₂V₂

5.90 × 0.250 = M₂ × 2

1.475 = M₂ × 2

Divide both side by 2

M₂ = 1.475 / 2

M₂ = 0.74 M

Thus, the molarity of the diluted solution is 0.74 M

En la fermentación del alcohol, la levadura convierte la glucosa en etanol y dióxido de carbono:
C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)
Si reaccionan 5.97 g de glucosa y se recolectan 1.44 L de CO2 gaseoso, a 293 K y 0.984 atm, ¿cuál
es el rendimiento porcentual de la reacción

Answers

Answer:

88.9%

Explanation:

Primero convertimos 5.97 g de glucosa a moles, usando su masa molar:

5.97 g ÷ 180 g/mol = 0.0332 mol

Después calculamos la cantidad máxima de moles de CO₂ que se hubieran podido producir:

0.0332 mol C₆H₁₂O₆ * [tex]\frac{2molCO_2}{1molC_6H_{12}O_6}[/tex] = 0.0664 mol CO₂

Ahora calculamos los moles de CO₂ producidos, usando los datos de recolección dados y la ecuación PV=nRT:

0.984 atm * 1.44 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 Kn = 0.0590 mol

Finalmente calculamos el rendimiento porcentual:

0.0590 mol / 0.0664 mol * 100% = 88.9%

A beaker is filled to the 500 mL mark with alcohol. What increase in volume (in mL) does the beaker contain when the temperature changes from 5° C to 30° C? (Neglect the expansion of the beaker, evaporation of alcohol and absorption of water vapor by alcohol.) The volume coefficient of expansion γγ for alcohol = 1.12 x 10-4 K-1

Answers

Answer:

"1.4 mL" is the appropriate solution.

Explanation:

According to the question,

[tex]v_0=500[/tex][tex]\alpha =1.12\times 10^{-4}[/tex][tex]\Delta \epsilon = 25[/tex]

Now,

Increase in volume will be:

⇒ [tex]\Delta V = \alpha\times v_0\times \Delta \epsilon[/tex]

By putting the given values, we get

           [tex]=1.12\times 10^{-4}\times 500\times 25[/tex]

           [tex]=1.12\times 10^{-4}\times 12500[/tex]

           [tex]=1.4 \ mL[/tex]

A major component of gasoline is octane when octane is burned in air it chemically reacts with oxygen to produce carbon dioxide and water what mass of carbon dioxide is produced by the reaction of oxygen

Answers

gasoline is the chemical that is coming out of the air

Which equation obeys the law of conservation of
mass?

Answers

Answer:2C4H10+2C12+12O2 4CO2+CC14+H20

what type of properties change ina physical change? Give an example to support your answer?



pls quick who will give the answer first will get the brainliest​

Answers

Explanation:

We can observe some physical properties, such as density and color, without changing the physical state of the matter observed. Other physical properties, such as the melting temperature of iron or the freezing temperature of water, can only be observed as matter undergoes. A physical change physical change involves a change in physical properties. Examples of physical properties include melting, transition to a gas, change of strength, change of durability, changes to crystal form, textural change, shape, size, color, volume and density.

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