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1 year ago

How i the ma number of an atom calculated? (1 point)

1 answer:

4 0

Proton plus neutron is the correct answer. Protons and neutrons have a mass of 1 and electrons have a mass of 0. So in order to find the mass of an atom you need to add the number of protons and the number of neutrons.

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Is it possible for an electron in a hydrogen atom to emit a gamma ray by falling to a lower energylevel?

jeka94

No, its not possible. Let's go back to 4th grades' basic science.
"Energy can not be created or destroyed."
This applies to everything, even atoms!

Have a great day! ❤

7 0

3 years ago

Am I correct lots of points!

larisa [96]

Yes the answer is correct the plant uses the energy to grow and fertilize other plants

5 0

3 years ago

Read 2 more answers

How many grams of CaF2 would be needed to produce 1.12 moles of F2?

zlopas [31]

Step 1 : Write balanced chemical equation.

CaF₂ can be converted to F₂ in 2 steps. The reactions are mentioned below.

I] $CaF_{2} + H_{2} SO_{4} -------> 2HF + CaSO_{4}$

II] $2HF -------> H_{2} + F_{2}$

The final balanced equation for this reaction can be written as

$CaF_{2} + H_{2} SO_{4} ---------> CaSO_{4} + H_{2} + F_{2}$

Step 2: Find moles of CaF₂ Using balanced equation

We have 1.12 mol F₂

The mole ratio of CaF₂ and F₂ is 1:1

$1.12mol F_{2} * \frac{1molCaF_{2}}{1molF_{2}} = 1.12molCaF_{2}$

Step 3 : Calculate molar mass of CaF2.

Molar mass of CaF₂ can be calculated by adding atomic masses of Ca and F

Molar mass of CaF₂ = Ca + 2 (F)

Molar mass of CaF₂ = 40.08 + 18.998 = 78.08 g

Step 4 : Find grams of CaF₂

Grams of CaF₂ = $1.12molCaF_{2} *\frac{78.08gCaF_{2}}{1mol CaF_{2}}$

Grams of CaF₂ = 87.45 g

87.45 grams of CaF2 would be needed to produce 1.12 moles of F2.

4 0

3 years ago

Read 2 more answers

1. How are the valence electrons in lithium and sodium atoms different?

Eva8 [605]

Answer:

if you look at the number at the top of the element square you can find the amount of positive atoms wich is also the amount of negative atoms. so if you count the valence electrons then that would be you difference

explanation:

Valence electrons are the amount of negative atoms on the outermost shell.

6 0

3 years ago

Calculate the radius of tantalum (Ta) atom, given that Ta has a BCC crystal structure, a density of 16.6 g/cm, and an atomic wei

Ivahew [28]

Answer:

The radius of tantalum (Ta) atom is $R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm$

Explanation:

From the Body-centered cubic (BBC) crystal structure we know that a unit cell length a and atomic radius R are related through

$a=\frac{4R}{\sqrt{3} }$

So the volume of the unit cell $V_{c}$ is

$V_{c}= a^3=(\frac{4R}{\sqrt{3} } )^3=\frac{64\sqrt{3}R^3}{9}$

We can compute the theoretical density ρ through the following relationship

$\rho=\frac{nA}{V_{c}N_{a}}$

where

n = number of atoms associated with each unit cell

A = atomic weight

$V_{c}$ = volume of the unit cell

$N_{a}$ = Avogadro’s number ( $6.023 \times 10^{23}$ atoms/mol)

From the information given:

A = 180.9 g/mol

ρ = 16.6 g/cm^3

Since the crystal structure is BCC, n, the number of atoms per unit cell, is 2.

We can use the theoretical density ρ to find the radio R as follows:

$\rho=\frac{nA}{V_{c}N_{a}}\\\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}$

Solving for R

$\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}\\\frac{64\sqrt{3}R^3}{9}=\frac{nA}{\rho N_{a}}\\R^3=\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}} \\R=\sqrt[3]{\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}$

Substitution for the various parameters into above equation yields

$R=\sqrt[3]{\frac{2\cdot 180.9}{16.6\cdot 6.023 \times 10^{23}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}\\R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm$

7 0

3 years ago