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ladessa [460]
3 years ago
13

OK SO HERES HOW ITS GONNA GO. IMA ASK YOU GUYS ONE MORE TIME BC THIS IS LIKE THE 3RD TIME IVE WAISTED MY POINTS IN THE SAME QUES

TION. IGHT JUST TRY.
MY TEACHER PIS.S.ES ME OFF I NEED HELP I SWEAR.

Question:
Give the electron configuration for the calcium ion. (2points)

My Answer:
(AR)4s^2

0/2 points

Teachers response:
This is asking for the ion and that is the neutral configuration. Write out the whole thing.
^
|
|
Help on that plz
Chemistry
1 answer:
Ne4ueva [31]3 years ago
4 0

Answer:

Electronic Configuration of calcium ion: Ca²⁺

1s², 2s², 2p⁶, 3s², 3p⁶

Explanation:

Calcium belong to group IIA and known as alkaline earth metals.

  • it have two electrons in outer most orbital.
  • these outer two electrons are involve in bonding
  • calcium lose these two electron and form cation
  • cation have 2+ charge
  • calcium have total twenty electrons

Electronic Configuration of calcium atom: ²⁰Ca

1s², 2s², 2p⁶, 3s², 3p⁶, 4s²

When calcium ion is form it loses two electron and have 2+ charge on it.

and have total of 18 electrons

So,

Electronic Configuration of calcium ion: Ca²⁺

the last filled orbital will be 3p⁶

1s², 2s², 2p⁶, 3s², 3p⁶

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Answer:

ρ = 1.08 g/cm³

Explanation:

Step 1: Given data

Mass of the substance (m): 21.112 g

Volume of the substance (V): 19.5 cm³

Step 2: Calculate the density of the substance

The density (ρ) of a substance is equal to its mass divided by its volume.

ρ = m / V

ρ = 21.112 g / 19.5 cm³

ρ = 1.08 g/cm³

The density of the substance is 1.08 g/cm³.

6 0
3 years ago
In a two-step synthesis, C6H11Br is converted into C6H12O. From the structure of the product, molecular formula of the starting
tamaranim1 [39]

Answer:

See explanation below

Explanation:

The question is incomplete. However in picture 1, you have the starting materials and the structure of the product, which you miss in this part.

Now, in picture 2, you have the starting reactant and the product, and the mechanism that is taking place here.

First, all what we have here is an acid  base reaction. In the first step, we are using the acid medium to convert the reactant into an alcohol. The bromine there, is not leaving the molecule yet, because it's neccesary for the next step. The starting reactant is an alkene, in that way, we can convert the reactant in the first step into a secondary alcohol. In other words, the first reaction is a alkene hydration.

In the second step, we use a strong base. You may say this is a strong nucleophile and will do a Sn2 reaction to form another alcohol there, but it's not the case, because, before any kind of reaction happens, the priority here is always the acid base, so the base will react with the acidic hydrogen. In this case, it will substract an hydrogen from the OH. When this happens, the lone pair will do an auto condensation here, and attacks the bromine in the molecule. In this way, the molecule will become a cyclomolecule, and that way it form the final product.

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8 0
3 years ago
If your front yard is 16.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1250 new snowflakes every minu
Free_Kalibri [48]
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3 years ago
The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p
saul85 [17]

Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

4 0
2 years ago
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Dahasolnce [82]
False. Chemical products are on the right side.
6 0
2 years ago
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