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How many grams of molybdenum(Mo) are in 2.68E24 atoms of Mo?

loris [4]

<h3>Answer:</h3>

427 g Mo

<h3>General Formulas and Concepts:</h3>

<u>Chemistry</u>

<u>Atomic Structure</u>

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

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<h3>Explanation:</h3>

<u>Step 1: Define</u>

2.68E24 atoms Mo or 2.68 × 10²⁴ atoms Mo

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of Mo - 95.94 g/mol

<u>Step 3: Convert</u>

<u /> $2.68 \cdot 10^{24} \ atoms \ Mo(\frac{1 \ mol \ Mo}{6.022 \cdot 10^{23} \ atoms \ Mo} )(\frac{95.94 \ g \ Mo}{1 \ mol \ Mo} )$ = 426.966 g Mo

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

426.966 g Mo ≈ 427 g Mo

3 0

3 years ago

The Brønsted-Lowry theory defines an acid as a(n)

Novosadov [1.4K]

Letra A. Doador de H+

6 0

4 years ago

A sulfur containing amino acid is

SVETLANKA909090 [29]

Cysteine and methionine contain sulfur.

6 0

3 years ago

If a temperature increase from 21.0 ∘c to 35.0 ∘c triples the rate constant for a reaction, what is the value of the activation

Airida [17]

Answer:

59.077 kJ/mol.

Explanation:

From Arrhenius law: <em>K = Ae(-Ea/RT)</em>

where, K is the rate constant of the reaction.

A is the Arrhenius factor.

Ea is the activation energy.

R is the general gas constant.

T is the temperature.

At different temperatures:

<em>ln(k₂/k₁) = Ea/R [(T₂-T₁)/(T₁T₂)]</em>

k₂ = 3k₁ , Ea = ??? J/mol, R = 8.314 J/mol.K, T₁ = 294.0 K, T₂ = 308.0 K.

ln(3k₁/k₁) = (Ea / 8.314 J/mol.K) [(308.0 K - 294.0 K) / (294.0 K x 308.0 K)]

∴ ln(3) = 1.859 x 10⁻⁵ Ea

∴ Ea = ln(3) / (1.859 x 10⁻⁵) = 59.077 kJ/mol.

4 0

3 years ago

A light source of wavelength, l, illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1 eV. A second l

madreJ [45]

Explanation:

From first source, kinetic energy ( $K.E_{1}$ ) ejected is 1 eV and wavelength of light is $\lambda$ .

From second source, kinetic energy ( $K.E_{2}$ ) ejected is 4 eV and wavelength of light is $\frac{\lambda}{2}$ .

Relation between work function, wavelength, and kinetic energy is as follows.

K.E = $\frac{hc}{\lambda} - \phi$

where, h = Plank's constant = $6.63 \times 10^{-34}$ J.s

c = speed of light = $3 \times 10^{8}$ m/s

Also, it is known that 1 eV = $1.6 \times 10^{-19}$ J

Therefore, substituting the values in the above formula as follows.

From first source,

$K.E_{1}$ = $\frac{hc}{\lambda} - \phi$

1 eV = $1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi$

$1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (1)

From second source,

$K.E_{2}$ = $\frac{hc}{\lambda} - \phi$

$4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi$

$6.4 \times 10^{-19} J = \frac{2 \times 1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (2)

Now, divide equation (2) by 2. Therefore, it will become

${6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}$

$3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}$ ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

$1.6 \times 10^{-19} = \frac{\phi}{2}$

$\phi$ = $3.2 \times 10^{-19}$

= 2 eV

Thus, we can conclude that work function of the metal is 2 eV.

3 0

4 years ago

That big lorry hit the new car.change to passive voice​

That big lorry hit the new car.change to passive voice