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FromTheMoon [43]
2 years ago
8

NEED HELP!!!!!!

Chemistry
2 answers:
Mkey [24]2 years ago
6 0
Chlorine.

If you search Google images for "Aufbau principle periodic table," you'll find some handy diagrams that will make it much easier to determine an element based on its electron configuration. Determine the number of electrons in the last part of the configuration (in this case, 5), locate that group on the Aufbau periodic table diagram, then count that number from left to right within that group. In this case, within the "3p" portion of the Periodic table, count to 5 and you'll find Chlorine as the answer.
Alex73 [517]2 years ago
4 0

Answer:

chlorine

Explanation:

You might be interested in
How many grams of sodium metal are needed to react completely with 25.8 liters of chlorine gas at 293 Kelvin and 1.30 atmosphere
Kitty [74]

I think it takes about 278g but I'm not 100% sure


5 0
3 years ago
When you combine 50.0 mL of 0.100 M AgNO3 with 50.0 mL of 0.100 M HCl in a coffee-cup calorimeter, the temperature changes from
RideAnS [48]

Answer : The enthalpy of reaction (\Delta H_{rxn}) is, 67.716 KJ/mole

Explanation :

First we have to calculate the moles of AgNO_3 and HCl.

\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole

\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole

Now we have to calculate the moles of AgCl formed.

The balanced chemical reaction will be,

AgNO_3(aq)+HCl(aq)\rightarrow AgCl(s)+HNO_3(aq)

As, 1 mole of AgNO_3 react with 1 mole of HCl to give 1 mole of AgCl

So, 0.005 mole of AgNO_3 react with 0.005 mole of HCl to give 1 mole of AgCl

The moles of AgCl formed  = 0.005 mole

Total volume of the solution = 50.0 ml + 50.0 ml = 100.0 ml

Now we have to calculate the mass of solution.

Mass of the solution = Density of the solution × Volume of the solution

Mass of the solution = 1.00 g/ml × 100.0 ml = 100 g

Now we have to calculate the heat.

q=m\times C\Delta T=m\times C \times (T_2-T_1)

where,

q = heat

C = specific heat capacity = 4.18J/g^oC

m = mass = 100 g

T_2 = final temperature = 24.21^oC

T_1 = initial temperature = 23.40^oC

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

q=100g\times (4.18J/g^oC)\times (24.21-23.40)^oC

q=338.58J

Now  we have to calculate the enthalpy of the reaction.

\Delta H_{rxn}=\frac{q}{n}

where,

\Delta H_{rxn} = enthalpy of reaction = ?

q = heat of reaction = 338.58 J

n = moles of reaction = 0.005 mole

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

\Delta H_{rxn}=\frac{338.58J}{0.005mole}=6771.6J/mole=67.716KJ/mole

Conversion used : (1 KJ = 1000 J)

Therefore, the enthalpy of reaction (\Delta H_{rxn}) is, 67.716 KJ/mole

4 0
3 years ago
The solubility of KCl is 3.7 M at 20 °C. Two beakers each contain 100. mL of saturated KCl solution: 100. mL of 4.0 M HCl is add
JulijaS [17]

Answer:

a)The Ksp was found to be equal to 13.69

Explanation:

Terminology

Qsp of a dissolving ionic solid — is the solubility product of the concentration of ions in solution.

Ksp however, is the solubility product of the concentration of ions in solution at EQUILIBRIUM with the dissolving ionic solid.

Note that if Qsp > Ksp , the solid at a certain temperature, will precipitate and form solid. That means the equilibrium will shift to the left in order to attain or reach equilibrium (Ksp).

Step-by-step solution:

To solve this: 

#./ Substitute the molar solubility of KCl as given into the ion-product equation to find the Ksp of KCl.

#./ Find the total concentration of ionic chloride in each beaker after the addition of HCl. We pay attention to the amount moles present at the beginning and the moles added.

#./ Find the Qsp value to to know if Ksp is exceeded. If Qsp < Ksp, nothing will precipitate.

a) The equation of solubility equilibrium for KCL is thus;

KCL_(s) ---> K+(aq) + Cl- (aq)

The solubility of KCl given is 3.7 M.

Ksp= [K+][Cl-] = (3.7)(3.7) =13.69

The Ksp was found to be equal to 14.

In pure water KCl

Ksp =13.69 KCl =[K+][Cl-]

Let x= molar solubility [K+],/[Cl-] :. × , x

Ksp =13.69 = [K+][Cl-] = (x)(x) = x²

x= √ 13.69 = 3.7 M moles of KCl requires to make 100mL saturated solutio

37M moles/L

The Ksp was found to be equal to 14.

4.0 M HCl = KCl =[K+][Cl-]

Let y= molar solubility :. y, y+4

Ksp =13.69= [K+][Cl-] = (y)(y*+4)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (y)(y*+4)= y(4)

13.69=4y:. y= 3.42 moles/100mL

y= 34.2moles/L

8 M HCl = KCl =[K+][Cl-]

Let b= molar solubility :. B, b+8

Ksp =13.69= [K+][Cl-] = (b)(b*+8)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (b)(b*+8)= b(8)

13.69=8b:. b= 1.71 moles/100mL

17.1 moles/L

Therefore in a solution with a common ion, the solubility of the compound reduces dramatically.

8 0
3 years ago
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Help me outttttttttttt t t t. T t
Alla [95]

Answer:

the open bag allowed the gas from the reaction to escape

8 0
3 years ago
Is na20 a stable compound, or an unstable compound?
maria [59]

Answer:

unstable

Explanation:

3 0
2 years ago
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