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jeka57 [31]
3 years ago
13

One mole of beryllium (be) contains the same number of atoms as 22.99 g of sodium. There are about 2.007×1023 atoms in 9.0 g al.

One mole of calcium (ca) weighs about as much as two moles of neon (ne). 11.495 g of sodium contains 6.022×1023 atoms.
Chemistry
1 answer:
Hoochie [10]3 years ago
4 0

1) the number of atoms (or molecules or ions) in one mole of any substance is same= 6.023 X 10^23

Thus as mass of one mole of sodium = 22.99 grams, this is equal to molar mass of sodium

moles of sodium = mass / molar mass = 22.99/22.99 = 1

Thus number of atoms in Be will be same the number of atoms of 22.99g of sodium

2) The mass of aluminium is given = 9g

atomic mass of Al = 27

Moles of Al =  mass / molar mass = 9/ 27 = 1/3

The atoms of Al in one mole = 6.023 X 10^23

Hence atoms of Al in 1/3 moles of Al = 6.023 X 10^23 / 3 =   2.007 X 10^23

So given statement is true

3) The mass of one mole of Ca = 40grams

The mass of one mole of Neon = 20grams

So yes one mole of Ca will weigh equal to two moles of Ne

4) The atomic mass of sodium = 23 g / mole

Thus 23 grams of sodium contains 6.022×1023 atoms and not  11.495 g of sodium

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The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W
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0.0010 mol·L⁻¹s⁻¹  

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\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}

You are cutting each concentration in half, so

\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}

Then,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}

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5 0
2 years ago
Read 2 more answers
Give the n and l values and the number of orbitals for sublevel 5g.
Pepsi [2]

The n and l values and the number of orbitals for sublevel 5g is :

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1) Principal quantum number , n

2) Angular quantum number , l

3) Magnetic quantum number , ml

4) spin quantum number , ms

For 5g shell, n = 5

subshell g , l = 4     ....0 - s , 1 - p , 2 - d, 3 - f, 4 -g

number of orbitals in subshell = (2l + 1)  ( 2×4 + 1) = 9

Thus,  The n and l values and the number of orbitals for sublevel 5g is :

5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.

To learn more about quantum numbers here

brainly.com/question/14650894

#SPJ1

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