Ask question

Ask question

All categories

3 years ago

10

Calculate the atomic weight of rubidium, which has two isotopes with the following properties: 85^Rb (84.91 amu, 72.71% natural

occurrence) and 87^Rb (86.91 amu, 27.83% natural occurrence). Round your answer to 4 significant figures

1 answer:

liraira [26]3 years ago

3 0

Answer:

The atomic weight of Rubidium (Rb) is 85.91 amu, using 4 significant figures.

Explanation:

The average atomic mass of Rubidium (Rb) is composed by the mass of the two given isotopes 85^Rb and 87^Rb but each isotope has a different abundance in nature expressed by percentage (natural occurrence).

Therefore, we need to calculate how much of this mass with its abundance percentage contribute to the average atomic mass for each isotope.

First step: Calculate the contribution in mass using the natural occurrence percentage.

85^Rb = 84.91 × $\frac{72.71}{100}$ = 61.73 amu

87^Rb = 86.91 × $\frac{27.83}{100}$ = 24.18 amu

The answers for each isotope are given using a typical percentage formula.

Second step: Add the two mass values founded.

24.18 amu + 61.73 amu = 85.91 amu

Third Step: Check the significant figures

The answer is 85.91 amu and it has 4 significant figures.

You might be interested in

In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for

inessss [21]

Answer:

The standard cell potential of the reaction is 0.78 Volts.

Explanation:

$Cu^{2+}(aq)+Fe(s)\rightarrow Cu(s)+Fe^{2+}(aq)$

Reduction at cathode :

$Cu^+(aq)+2e^-\rightarrow Cu(s)$

Reduction potential of $Cu^{2+}$ to Cu= $E^o_{1}=0.34 V$

Oxidation at anode:

$Fe(s)\rightarrow Fe^{2+}(aq)+2e^-$

Reduction potential of $Fe^{2+}$ to Fe= $E^o_{2}=-0.44 V$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.34V -(-0.44 V)=0.78 V$

The standard cell potential of the reaction is 0.78 Volts.

3 0

3 years ago

The half-life of carbon-14 is 5,730 years. how old is a sample that is 75% daughter isotope and 25% parent isotope?

Hitman42 [59]

Half-life it tells you about the amount of time needed that half of the quantity of an isotope to disintegrate.

For carbon-14, assuming that the daughter isotope is a stable one and does not disintegrate further, you have:

<u>parent isotope</u> <u>daughter isotope</u> <u>years</u>

100% 0% 0

50% 50% 5,730

25% 75% 11,460

5 0

3 years ago

Which substance will require more energy to heat if all the samples have the same mass? A chart with two columns and nine rows.

Paraphin [41]

Here we have to identify the sample which need more energy to heat the sample 1 degree Celsius.

Among the given elements magnesium will require more energy than the others to heat.

As per the definition of specific heat of a compound, the amount of heat required to increase the temperature of the material 1 degree Celsius is the specific heat of the material.

The given data are-

substance specific heat

Lead 0.129

Tin 0.21

Silver 0.235

Iron 0.449

Calcium 0.647

Granite 0.803

Aluminium 0.897

Magnesium 1.023

From the given data lead, magnesium, iron and aluminium have the specific heat 0.129, 1.023, 0.449 and 0.897 respectively. Thus magnesium will require more energy than the others to heat.

6 0

3 years ago

Express the concentration of a 0.0320 M aqueous solution of fluoride, F−, in mass percentage and in parts per million (ppm). Ass

denis23 [38]

Answer:

607 ppm

Explanation:

In this case we can start with the <u>ppm formula</u>:

$ppm=\frac{mg~of~solute}{Litters~of~solution}$

If we have a solution of <u>0.0320 M</u>, we can say that in 1 L we have 0.032 mol of $F^-$ , because the molarity formula is:

$M=\frac{mol}{L}$

In other words:

$0.0320~M=\frac{mol}{1~L}$

$mol=0.032~M*1~L=0.032~mol$

$1~L~of~Solution=0.0320~mol~of~solute$

If we use the <u>atomic mass</u> of $F$ (19 g/mol) we can convert from mol to g:

$0.0320~mol~F^-\frac{19~g~F^-}{1~mol~F^-}~=~0.607~g$

Now we can <u>convert from g to mg</u> (1 g= 1000 mg), so:

$0.607~g\frac{1000~mg}{1~g}=607~mg$

Finally we can <u>divide by 1 L</u> to find the ppm:

$ppm=\frac{607~mg}{1~L}=~607~ppm$

<u>We will have a concentration of 607 ppm.</u>

I hope it helps!

4 0

4 years ago

Hybridization of NH3

nadya68 [22]

Answer:

Sp3 hybridization

Explanation:

The NH3 molecule, which consists of one lone pairs and three bond pair of electron on its valance shell due to lone pair bond pair repulsion makes bond angle of 107.5°resulting distorted tetrahedral geometry.

Hybridization =no. of bond pair +lone pair=3+1=4=sp3 hybridization

6 0

3 years ago