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3 years ago

What is electron configuration of a Mn atom in the ground state

2 answers:

5 0

Answer (AR)
Explanation: mn has an atomic number of 25 . Simplify use this information to obtain its electronic configuration . MN 1s2 2s2 2p6 3s2 3p6 4s2 3d5

dexar [7]3 years ago

5 0

What is the electron configuration of a Mn atom in the ground state

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What elements are found in the compound represented in the diagram

DedPeter [7]

hydrogen and carbon, hope that helped

3 0

4 years ago

The highest pressure ever produced in a laboratory setting was about 2.0 x 10 6 atm. If we have a

tigry1 [53]

Answer:

Volume = 72.7272

Explanation:

if only pressure and volume change, then we can do some simple math to find the answer.

2 x 10 ^ 6 times 1 x 10 ^-5 = 20

pressure and volume must equal 20

20 = 0.275 x volume

20 / 0.275 = new volume

I don't remember significant digits but your volume is 72.72727272 just repeated

8 0

3 years ago

How many liters of carbon dioxide will be produced at STP if 3.56 g calcium carbonate reacts completely with carbon dioxide? CaC

sp2606 [1]

Answer:

V = 0.798 L

Explanation:

Hello there!

In this case, for this gas stoichiometry problem, we first need to compute the moles of carbon dioxide via stoichiometry and the molar mass of starting calcium carbonate:

$3.56gCaCO_3*\frac{1molCaCO_3}{100gCaCO_3} *\frac{1molCO_2}{1molCaCO_3} =0.0356molCO_2$

Next, we use the ideal gas equation for computing the volume, by bearing to mind that the STP conditions stand for a pressure of 1 atm and a temperature of 273.15 K:

$PV=nRT\\\\V=\frac{nRT}{P}\\\\V=\frac{0.0356mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm} \\\\V=0.798L$

Best regards!

4 0

3 years ago

What were the limitations of Newlands Law of Octaves.?<br><br> No spam ❌❌<br> or else (-_-メ)

liubo4ka [24]

Answer:

The major limitations of Newlands' law of octaves were : (i) It was applicable to only lighter elements having atomic masses upto 40 u, i.e., upto calcium. After calcium, the first and the eighth element did not have similar properties

5 0

3 years ago

Read 2 more answers

How many moles in 4.93 x 10E23 atoms of silver?

leva [86]

<h3>Answer:</h3>

0.819 mol Ag

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

4.93 × 10²³ atoms Ag

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

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 4.93 \cdot 10^{23} \ atoms \ Ag(\frac{1 \ mol \ Ag}{6.022 \cdot 10^{23} \ atoms \ Ag})$
Divide: $\displaystyle 0.818665 \ mol \ Ag$

<u>Step 4: Check</u>

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

0.818665 mol Ag ≈ 0.819 mol Ag

8 0

3 years ago