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Elina [12.6K]
2 years ago
5

What is the electron configuration for 1s2 2s2 2p6 3s2 3p6​

Chemistry
1 answer:
Sliva [168]2 years ago
7 0
This is going to lead you to the element Argon. Is this what you are asking?
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What is the total ionic equation for cuso4+2naoh>>cu(oh)2+na2so4
iVinArrow [24]

Answer:

CuSO4 + 2 NaOH = Cu(OH)2 + Na2SO4

3 0
2 years ago
Which type of reaction occurs in the following equation?
Tems11 [23]
The  type  of reaction  in  above reaction is refered  to  as a disproportionation  reaction .It sometimes  used  to describe  desymmetrizing  reaction.in the equation  above compounds  of  intermediate  oxidation  state  are  converted  to  two  different  compound,  that  is Clo3  ions  and  2cl-
3 0
3 years ago
Calculate the value of the equilibrium constant, Kc , for the equilibrium shown below, if 0.124 moles of NO, 0.0240 mole of H2,
sdas [7]

the reaction is

2NO(g) + 2H2(g) <—> N2(g) + 2H2O (g)

Kc = [N2] [ H2O]^2 / [NO]^2 [ H2]^2

Given

moles of NO = 0.124 therefore [NO] = moles /volume = 0.124 /2 = 0.062

moles of H2 = 0.0240 , therefore [H2] = moles / volume = 0.0240 / 2 = 0.012

moles of N2 = 0.0380 , therefore [N2] = moles / volume = 0.0380 / 2 = 0.019

moles of H2O  = 0.0276 , therefore [H2O] = moles / volume = 0.0276 / 2 = 0.0138

Kc = (0.019) ( 0.0138)^2 / (0.062)^2 ( 0.012)^2 = 6.54



4 0
2 years ago
Identify an alkene and carboxylic acid using primary observations
ehidna [41]

Answer:

The general formula for the carboxylic acids is C nH 2n+1COOH (where n is the number of carbon atoms in the molecule, minus 1).

Explanation:

<em>Hope </em><em>it </em><em>helps </em><em>u </em>

FOLLOW MY ACCOUNT PLS PLS

4 0
3 years ago
A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C・g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('
polet [3.4K]

Answer:

a=5.65cm

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

Q_{Pt}=-Q_{Deu}

We can represent the heats in terms of mass, heat capacities and temperatures:

m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})

Thus, we solve for the mass of platinum:

m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g

Next, by using the density of platinum we compute the volume:

V_{Pt}=\frac{3860.4g}{21.45g/cm^3}\\ \\V_{Pt}=180cm^3

Which computed in terms of the edge length is:

V=a^3

Therefore, the edge length turns out:

a=\sqrt[3]{180cm^3}\\ \\a=5.65cm

Best regards.

6 0
3 years ago
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