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3 years ago

Calculate the number of atoms in 52g of He.

Chemistry

1 answer:

MissTica3 years ago

5 0

Answer:

78.3 × 10²³ atoms of helium are present in 52 g.

Explanation:

Given data:

Mass of He = 52 g

Number of atoms = ?

Solution:

First of all we will calculate the number of moles of He

Number of moles = mass /molar mass

Number of moles = 52 g/ 4 g/mol

Number of moles = 13 mol

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

1 mole = 6.022 × 10²³ atoms of helium

13 mol × 6.022 × 10²³ atoms of helium / 1 mole

78.3 × 10²³ atoms of helium

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When radium-226 (atomic number-88) decays by emitting an alpha particle it becomes _____.

topjm [15]

Answer:

d. Radon-222

Explanation:

²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He

Alpha particle is a helium nucleus with mass number 4 and atomic number 2. According to the law of conversation of mass, the sum of the mass number and atomic number must be equal on both side of the reaction.

Since the mass number of Ra is 226 and that of He is 4. The mass number of the unknown element must be 226 - 4 = 222.

Since the atomic number of Ra is 88 and that of He is 2. The atomic number of the unknown element must be 88 - 2 = 86.

Now looking in the periodic table Radon is the only element with atomic number 86.

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3 years ago

In a redox reaction, what role does the reducing agent play?

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<span>So the oxidizing agent will receive electrons from the reducing agent and the oxidation agent will take electrons from the reducing agent.</span>

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3 years ago

Read 2 more answers

Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of a conservative pollutant, is released int

hjlf

Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.

Explanation:

The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.

Million Gallons per day $1 MGD = 3785411.8 litre/day = 3785411.8 L/d$

$F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d$

We have one flow of wastewater released into a stream.

First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.

Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:

$C_f = \frac{F1*C1 +F2*C2}{F1 +F2}$

Replacing every value in L/d and mg/L

$C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L$

a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.

Finally, we have to calculate the pounds of substance per day (Mp).

We have the total flow F3 = F1 + F2 and the final concentration $C_f$ . It is required to calculate per day, let's take a time of t = 1 day.

$F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg$

After that, mg are converted to pounds.

$M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb$

b) A total of 137.6 pounds pass a given spot downstream per day.

4 0

3 years ago

A compound consisting of B, N, and H undergoes elemental analysis. The % composition by mass is found to be 40.28% B, 52.20% N,

swat32

Empirical formula is the simplest ratio of components making up a compound.

The percentage composition of each element has been given

therefore the mass present of each element in 100 g of compound is

B N H

mass 40.28 g 52.20 g 7.53 g

number of moles

40.28 g / 11 g/mol 52.20 g / 14 g/mol 7.53 g / 1 g/mol

= 3.662 mol = 3.729 mol = 7.53 mol

divide the number of moles by the least number of moles, that is 3.662

3.662 / 3.662 3.729 / 3.662 7.53 / 3.662

= 1.000 = 1.018 = 2.056

the ratio of the elements after rounding off to the nearest whole number is

B : N : H = 1 : 1 : 2

therefore empirical formula for the compound is B₁N₁H₂

that can be written as BNH₂

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3 years ago

If a chemist has 2.3 moles of arsenic (as) that has a molar mass of 74.92 g/mol, which would be the best way to find out how man

mr Goodwill [35]

The best way to determine the number of atoms of arsenic in the sample will be to multiply 2.3 by Avagadro's number.
This is because Avagadro's number is the number of particles one mole of any substance has, and its value is 6.02 x 10²³
If the number of moles of a substance are known, then multiplying by Avagadro's number will give the number of particles. In this case, this is 1.38 x 10²⁴.

3 0

3 years ago

Calculate the number of atoms in 52g of He.​

Calculate the number of atoms in 52g of He.