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Zinaida [17]
3 years ago
11

Consider a liquid in a piston apparatus that is at equilibrium with the vapor phase. in which direction should the piston be mov

ed to decrease the number of molecules in the liquid phase?
Chemistry
1 answer:
EastWind [94]3 years ago
7 0
<span>the piston should be in the upmost position to allow room for material to expand, thus be in vapor state, when a gas is compressed, it liquefies, and when dealing with a piston, much like in an auto combustion engine, the piston moves up and allows oxygen to mix with vapor and explode, thus powering further movement of pistons.</span>
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. For the reaction 2HNO3 + Mg(OH), → Mg(NO3)2 + 2H20, how many grams of magnesium nitrate are produced from 8.00 mol of nitric a
GREYUIT [131]

Mass of Magnesium nitrate produced : 593.2 g

<h3>Further explanation</h3>

The reaction equation is the chemical formula of reagents and product substances

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

2HNO₃ + Mg(OH)₂ → Mg(NO₃)₂ + 2H₂0

mol HNO₃ = 8

From the equation, mol ratio of HNO₃ : Mg(NO₃)₂ = 2 : 1, so mol Mg(NO₃)₂ :

\tt \dfrac{1}{2}\times 8=4~moles

Mass Mg(NO₃)₂(MW=148,3 g/mol) :

\tt mass=mol\times MW\\\\mass=4\times 148,3 g/mol\\\\mass=593.2~g

6 0
3 years ago
Give the value of the quantum number ℓ, if one exists, for a hydrogen atom whose orbital angular momentum has a magnitude of 6√(
MakcuM [25]

Answer:

For this angular momentum, no quantum number exist

Explanation:

From the question we are told that

   The magnitude of the angular momentum is  L  = 6\sqrt{\frac{h}{2 \pi} }

The generally formula for Orbital angular momentum is mathematically represented as

           L  = \sqrt{(l * (l + 1)) } *  \frac{h}{2 \pi}

Where l is the quantum number

now  

We can look at the given angular momentum in this form as

      L  = 6\sqrt{\frac{h}{2 \pi} }    =  \sqrt{36}  * \sqrt{\frac{h}{2 \pi}} }

comparing this equation to the generally equation for Orbital angular momentum

     We see that there is no quantum number that would satisfy this equation

5 0
3 years ago
The job of the human heart is to circulate blood through all of the body's arteries, capillaries, and veins. The job of the huma
weeeeeb [17]
B
All cells new oxygen and blood is a way for them to get it 

3 0
3 years ago
Read 2 more answers
239<br> 239<br> -<br> 93Np<br> 94Pu + c
seraphim [82]

Answer:4>34.1>56

Explanation:ez

5 0
2 years ago
5g of a mixture of KOH and KCl with water form a solution of 250mL. We have 25ml of this solution and we mix it with 14,3mL of H
cricket20 [7]
We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl

Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.

To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH

Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572.  That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH

Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH

Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH

I hope this helps.

7 0
3 years ago
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