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DedPeter [7]
2 years ago
10

Why does shampoo remove hair grese for chemistry acids and alkali

Chemistry
1 answer:
djverab [1.8K]2 years ago
6 0
Because it’s shampoo
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Identify the substances that will appear in the equilibrium constant expression for the equation: 2Ag+(aq)+Zn(s)<->Zn2+(aq
Simora [160]

Hey there!


We Know that:



 2 Ag⁺(aq) + Zn(s) <-> Zn²⁺(aq)+2 Ag(s)


The equilibrium expression for the reaction is:



Kc =  [ Zn⁺² ]  /  [Ag⁺ ]²


Hope that helps!

8 0
3 years ago
Read 2 more answers
Which group tends not to form ions or react?
Anastaziya [24]
Group 8 elements. They are unreactive and stable
3 0
3 years ago
Whats is causing the ice block to melt
slavikrds [6]

Answer:

see explanation

Explanation:

Ice usually has tightly packed molecules at a low temperature. When it comes in contact with a higher temperature or room temperature, the ice molecules gain energy and the molecular tension increases which causes the state to change to liquid. Therefore, a <u>high temperature</u> causes an ice block to melt.

5 0
3 years ago
If 2.4 g of n2 gas has a volume of 0.40 l and a pressure of 6.6 atm , what is its kelvin temperature?
Nesterboy [21]

To solve for the absolute temperature, we assume ideal gas behaviour so that we use the equation:

PV = nRT

or T = PV / nR

 

So calculating:

T = [6.6 atm * 0.40 L] / [(2.4g / 28g/mol) * 0.08205746 L atm / mol K]

<span>T = 375.35 K</span>

8 0
3 years ago
A closed vessel system of volume 2.5 L contains a mixture of neon and fluorine. The total pressure is 3.32 atm at 0.0°C. When th
gayaneshka [121]

Answer:

moles Ne = 0.154 mol

moles F₂ = 0.217 mol

Explanation:

Step 1: Data given

Volume of the vessel system = 2.5 L

Total pressure = 3.32 atm at 0.0 °C

The mixture is heated to 15.0 °C

The entropy of the mixture increases by 0.345 J/K

The heat capacity of monoatomic gas = 3/2R and that for a diatomic gas = 5/2R

Step 2: Define the gas

Neon is a monoatomic gas, composed of Ne atoms

 ⇒ Cv(Ne) ≅ (3/2)R

Fluorine is a diatomic gas, composed of F₂ molecules.  

⇒ Cv(F₂) ≅ (5/2)R

Step 3: Calculate moles of gas

p*V = n*R*T

⇒ with p = the total pressure = 3.32 atm

⇒ with V = the total volume = 2.5 L

⇒ with n = the number of moles of gas

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 273.15 Kelvin

n(total) = p*V/RT = (3.32 atm*2.5 L)/(0.08206 L*atm/mol•K*273.15) = 0.3703 mol

Step 4: Calculate moles of Ne and F2

For one mole heated at constant volume,  

∆S = Cv*ln(288.15/273.15) = 0.05346*Cv

⇒ ∆S for 0.3703 mol,  

∆S = (0.3703 mol)(0.05346)Cv = 0.345 J/K

 ⇒ Cv = 17.43 J/mol*K for the Ne/F₂ mixture.

For pure Ne, Cv = (3/2)R = 1.5*8.314 J/mol*K = 12.471 J/mol*K

For pure F₂, Cv = (5/2)R = 2.5 * 8.314 J/mol*K = 20.785 J/mol*K

if X is the mole fraction of Ne, we can find X by:

17.43 J/mol*K = X* 12.471 J/mol*K + (1 – X) * 20.785 J/mol*K

 ⇒ 20.875 – 8.314 * X = 17.43

X = 0.415 , 1 – X = 0.585

moles Ne = (0.415)(0.3703 mol) = 0.154 mol

moles F₂ = (0.585)(0.3703 mol) = 0.217 mol

4 0
3 years ago
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