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3 years ago

How are a substance and a solution similar? how are they different?

1 answer:

5 0

A substance and a solution have both uniform composition throughout. A substance is a matter that is uniform in composition and can be an element or compound. A solution is another name for homogeneous which is a mixture. The difference of substances and solution, a substance has a definite composition and a solution has a variable composition. It is also uniform in composition, but a solution is made by combining two or more substances.

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Hydrogen is the only atom that does not have ____ electrons in the 1st energy level/shell.

dalvyx [7]

Hydrogen is the only atom that does not have neutron electrons in the first energy level/shell.

4 0

3 years ago

Which statement about the greenhouse effect is NOT true?

Archy [21]

<span> Greenhouse gases were not historically present in the atmosphere.</span>

7 0

2 years ago

For the reaction 2 H2S(g) D 2 H2 (g) + S2 (g), Kp = 1.5 × 10−5 at 800.0°C. If the initial partial pressures of H2 and S2 in a cl

Usimov [2.4K]

Answer: The approximate equilibrium partial pressure of $H_2S$ is 3.92 atm

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

$2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)$

$K_p=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}$

$1.5\times 10^{-5}=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}$

On reversing the reaction:

$2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)$

initial pressure 4.00atm 2.00 atm 0

eqm (4.00-2x)atm (2.00-x) atm 2x atm

$K_p=\frac{[H_2S]^2}{[H_2]^2\times [S_2]}$

$K_p'=\frac{1}{K_p}=0.67\times 10^5$

$2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)$

$0.67\times 10^5=\frac{2x]^2}{[4.00-2x]^2\times [2.00-x]}$

$x=1.96$

$[H_2S]=2x=2\times 1.96=3.92 atm$

Thus approximate equilibrium partial pressure of $H_2S$ is 3.92 atm

3 0

3 years ago

What are the molality and mole fraction of solute in a 22.2 percent by mass aqueous solution of formic acid (HCOOH)?

hodyreva [135]

Calculate the mass of water used
that is
100-22.2=77.8g convert into Kg = 77.8/1000=0.0778Kg of water

then calculate the moles of HCOOH used
that is 22.2g/molar mass of HCOOH(1+12+16+16+1)=46
therefore the moles of HCOOH=22.2/46=0.48moles
the mole of water= 77.8/18(molar mass of water= 4.32moles

the molarity of HCOOH = 0.48mol/0.0778kg=6.17M

The mole ratio= moles of HCOOH divided by total moles
the total moles= 0.48+4.32=4.8moles
therefore the mole ratio= 0.48/4.8moles=0.1(the moles fraction of HCOOH)

7 0

3 years ago

The equilibrium constant is given for two of the reactions below. Determine the value of the missing equilibrium constant. 2A(g)

monitta

Answer:

The value of the missing equilibrium constant ( of the first equation) is 1.72

Explanation:

First equation: 2A + B ↔ A2B Kc = TO BE DETERMINED

⇒ The equilibrium expression for this equation is written as: [A2B]/[A]²[B]

Second equation: A2B + B ↔ A2B2 Kc= 16.4

⇒ The equilibrium expression is written as: [A2B2]/[A2B][B]

Third equation: 2A + 2B ↔ A2B2 Kc = 28.2

⇒ The equilibrium expression is written as: [A2B2]/ [A]²[B]²

If we add the first to the second equation

2A + B + B ↔ A2B2 the equilibrium constant Kc will be X(16.4)

But the sum of these 2 equations, is the same as the third equation ( 2A + 2B ↔ A2B2) with Kc = 28.2

So this means: 28.2 = X(16.4)

or X = 28.2/16.4

X = 1.72

with X = Kc of the first equation

The value of the missing equilibrium constant ( of the first equation) is 1.72

7 0

3 years ago