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2 years ago

Substance X is a compound containing 632mg of manganese and 368mg of oxygen. Substance X is shown

Chemistry

1 answer:

defon2 years ago

8 0

The empirical formula : MnO₂.

<h3>Further explanation</h3>

Given

632mg of manganese(Mn) = 0.632 g

368mg of oxygen(O) = 0.368 g

M Mn = 55

M O = 16

Required

The empirical formula

Solution

You didn't include the pictures, but the steps for finding the empirical formula are generally the same

Find mol(mass : atomic mass)

Mn : 0.632 : 55 = 0.0115

O : 0.368 : 16 =0.023

Divide by the smallest mol(Mn=0.0115)

Mn : O =

$\tt \dfrac{0.0115}{0.0115}\div \dfrac{0.023}{0.0115}=1\div 2$

The empirical formula : MnO₂

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Shifts in the rock layer locations cannot account for gaps in the rock record<br><br> true or false

vivado [14]

Answer: That would be false because it is the contact between two layers representing a gap in the geologic record, usually from the erosion of the layers which would normally be expected to appear.

Explanation:

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4 0

3 years ago

Ammonia can be prepared by the reaction of magnesium nitride with water. The products are ammonia and magnesium hydroxide. Write

deff fn [24]

Answer:

Mg3N2 +6H2O--> 2NH3 + 3Mg (OH)2

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2 years ago

Read 2 more answers

What is the number of the lowest energy level that contains an f sublevel?

forsale [732]

Answer: n = 4.

Explanation:

1) The main energy level is represented by the principal (main) quantum number, n. It can be 1, 2, 3, 4, 5,6, or 7. It is the period number in the periodic table.

2) The next quantum number is the Azimuthal quantum number (ℓ), also known as angular quantum number. It indicates the subshell (sublevel) or type of orbital.

This quantun number may be:

s, p, d, or f.

3) For n = 1, there is only one ℓ number: 0; which is the orbital s

For n = 2, there are two possible ℓ numbers: 0 and 1; which are the orbitals s and p.

For n = 3, there are three possible ℓ numbers: 0, 1, and 2, which are the orbitals s, p, and d:

For n = 4, there are four possible ℓ numbers:0, 1, 2, and 3; which are the orbitals s, p, d, and f.

So, the f sublevel appears first time when n = 4, i.e the number of the lowest energy level that contains an f sublevel is 4.

5 0

3 years ago

Question 3) A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in LiF. Calculate the pH of the solution after the addition of

Masja [62]

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

<u>Explanation:</u>

We have the chemical equation,

HF (aq)+NaOH(aq)->NaF(aq)+H2O

To find how many moles have been used in this

c= n/V=> n= c.V

nHF=0.250 M⋅1.5 L=0.375 moles HF

Simillarly

nF=0.250 M⋅1.5 L=0.375 moles F

nHF=0.375 moles - 0.250 moles=0.125 moles

nF=0.375 moles+0.250 moles=0.625 moles

[HF]=0.125 moles/1.5 L=0.0834 M

[F−]=0.625 moles/1.5 L=0.4167 M

To determine the problem using the Henderson - Hasselbalch equation

pH=pKa+log ([conjugate base/[weak acid])

Find the value of Ka

pKa=−log(Ka)

pH=−log(Ka) +log([F−]/[HF]

pH= -log(3.5 x 10 ^4)+log(0.4167 M/0.0834 M)

pH=-log(3.5 x 10 ^4)+log(4.996)

pH= -4.54+0.698

pH=-(-3.84)

pH=3.84

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

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3 years ago

Determine the expected diffraction angle for the first-order diffraction from the (111) set of planes for FCC nickel (Ni) when m

faust18 [17]

Answer:

56°

Explanation:

First calculate $a:$

$a=2 R \sqrt{2}=2(0.1246) \sqrt{2}=0.352 \mathrm{nm}$

The interplanar spacing can be calculated from:

$d_{111}=\frac{a}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{0.352}{\sqrt{3}}=0.203 \mathrm{nm}$

The diffraction angle is determined from:

$\sin \theta=\frac{n \lambda}{2 d_{111}}=\frac{1(0.1927)}{2(0.2035)}=0.476$

Solve for $\theta$

$\theta=\sin ^{-1}(0.476)=28^{\circ}$

The diffraction angle is:

$2 \theta=2\left(28^{\circ}\right)=56^{\circ}$

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3 years ago