All categories

2 years ago

How many atoms are present in a sample of Calcium (Ca) weighing 127.38g?

Chemistry

1 answer:

kari74 [83]2 years ago

3 0

Answer:

1.9139 × 10²⁴ atoms Ca

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

127.38 g Ca

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of Ca - 40.08 g/mol

Step 3: Convert

$127.38 \ g \ Ca(\frac{1 \ mol \ Ca}{40.08 \ g \ Ca} )(\frac{6.022 \cdot 10^{23} atoms \ Ca}{1 \ mol \ Ca} )$ = 1.91388 × 10²⁴ atoms Ca

Step 4: Check

We are given 5 sig figs. Follow sig fig rules and round.

1.91388 × 10²⁴ atoms Ca ≈ 1.9139 × 10²⁴ atoms Ca

You might be interested in

HS- is amphoteric; it can behave as either an acid or a base.

Tamiku [17]

Answer:

HS+Na=>NaS+1/2H2(here HS- acts as an acid)

HS-. + HCl=> H2S(g)+ Cl-(here HS- acts as a base)

7 0

3 years ago

How many moles of Ar gas are present in a container with a volume of 78.4 L at STP?

rusak2 [61]

<h3>Answer:</h3>

1 mole of a gas at STP occupies 22.4 L volume 

Now the volume is given =78.4 therefore,

No. of moles of gas = 78.4 ÷ 22.4 = 3.5 moles

I hope it helps you~

7 0

2 years ago

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle

Sedaia [141]

Question in incomplete, complete question is:

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle (electron) with a kinetic energy (Ek) of $4.71\times 10^{-15}J$ . What is the de Broglie wavelength of this electron (Ek = ½mv²)?

Answer:

$6.762\times 10^{-12} m$ is the de Broglie wavelength of this electron.

Explanation:

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

$\lambda=\frac{h}{\sqrt{2mE_k}}$

where,

= De-Broglie's wavelength = ?

h = Planck's constant = $6.624\times 10^{-34}Js$

m = mass of beta particle = $9.1094\times 10^{-31} kg$

$E_k$ = kinetic energy of the particle = $4.71\times 10^{-15}J$

Putting values in above equation, we get:

$\lambda =\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 9.1094\times 10^{-31} kg\times 4.71\times 10^{-15}J}}$

$\lambda = 6.762\times 10^{-12} m$

$6.762\times 10^{-12} m$ is the de Broglie wavelength of this electron.

3 0

3 years ago

Described the role of electrons in the formation of a covalent bond

My name is Ann [436]

Sharing of two electrons make a Covalent bond.

Explanation:

Attractions among the atoms bring them together. So the electrons from each of the atoms are attracted towards the nucleus of those two atoms, that “share” the electrons produces a covalent bond.

It is also named as molecular bond, a bond that entails the sharing of a pair of electrons among the atoms. When the atoms share the electrons among themselves, it produces a molecule, which is more stable than the atom.

If the attractions between the atoms are strong enough and if every atom has enough space for the electrons in its outermost energy level then there occurs covalent bonding. So electrons are very important in the covalent bond formation.

7 0

3 years ago

a certain compound was found to contain 54.0 g of carbon and 10.5 grams of hydrogen. its relative molecular mass is 86.0. find t

dexar [7]

Answer:

empirical formula = C3H7

molecular formula = C6H14

3 0

3 years ago