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ValentinkaMS [17]
2 years ago
11

Which of the following happens when a reaction reaches dynamic equilibrium in a closed system?

Chemistry
1 answer:
laiz [17]2 years ago
8 0

Answer: B

Explanation:

At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, and thus the concentrations of the reactants and products must be constant.

You might be interested in
What is an Atom?<br><br> What are the 5 most common elements that make up the human body?.
dalvyx [7]
An atom is the smallest unit of ordinary matter that forms a chemical element.

oxygen, carbon, hydrogen, nitrogen, calcium, and phosphorus.
6 0
2 years ago
In Lab 10 you make a stock solution of salicylic acid, and then four dilutions. The stock solution is made by diluting 5.00 ml o
Zigmanuir [339]

Answer:

Stock  solution =  1.25 *10^-3 M

Dilution 1 = 5*10^-4 M

Dilution 2= 3.75 * 10^-4 M

Dilution 3 = 2.5 *10^-4 M

Dilution 4 = 1.25 *10^-4 M

Explanation:

<u>Step 1:</u> Data given

The stock solution is made by diluting 5.00 ml of 1.250 x 10-2 M salicylic acid in 50.00 mL of solution.

<u>Step 2</u>: Calculate the concentration of the stock solution:

M1*V1 = M2*V2

⇒ with M1 = the initial concentration = 1.250 *10^-2 M

⇒ with V1 = 5 mL = 5*10^-3 L

⇒ with M2 = TO BE DETERMINED

⇒ with V2 = 50 mL = 50 *10^-3 L

M2 = (M1*V1)/V2

M2 = (1.250 *10^-2 * 5*10^-3 L) / 50 *10^-3

M2 = 0.00125 M = 1.25 *10^-3 M

<u>Step 3:</u> Calculate dilution 1

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 *10^-3)/(25*10^-3L)

M2 = 0.0005 M = 5*10^-4 M

<u>Step 4</u>: Calculate dilution 2

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 * 7.5*10^-3)/(25*10^-3)

M2 = 0.000375 M = 3.75 * 10^-4 M

<u>Step 5:</u> Calculate dilution 3

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 5*10^-3) /(25*10^-3)

M2 = 0.00025 M = 2.5 *10^-4 M

<u>Step 6</u>: Calculate dilution 4

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 2.5*10^-3)/(25*10^-3)

M2 = 0.000125 M = 1.25 *10^-4 M

5 0
3 years ago
EC
REY [17]

Answer:

a

Explanation:

this is easy its a

8 0
3 years ago
Calculate the molality of a 20.0% by mass ammonium sulfate (nh4)2so4 solution. the density of the solution is 1.117 g/ml.
olasank [31]
Hello!

We have the following data:

m1 (solute mass) = 20 % m/m
M1 (Molar mass of solute) (NH4)2 SO4 = ?
m2 (mass of the solvent) = ? (in Kg)

First we find the solute mass (m1), knowing that:

20% m/m = 20g/100mL

20 ------ 100 mL (0,1 L)
y g --------------- 1 L

y = 20/0,1 
y = 200 g --> m1 = 200 g

Let's find Solute's Molar Mass, let's see:

M1 of (Nh4)2SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

We must find the volume of the solvent and therefore its mass (m2), let us see:

d = 1,117 g/mL
m = 200 g
v (volumen of solute) = ?

d =  \dfrac{m}{V} \to V =  \dfrac{m}{d}

V =  \dfrac{200\:\diagup\!\!\!\!g}{1,117\:\diagup\!\!\!\!g/mL} \to V = 179\:mL\:(volumen\:of\:solute)

<span>The solvent volume will be:
</span>
1000 -179 => V = 821 mL (volumen of disolvent)

If: 1 mL = 1g

<span>Then the mass of the solvent is:
</span>
m2 (mass of the solvent) = 821 g → m2 (mass of the solvent) = 0,821 Kg

Now, we apply all the data found to the formula of Molality, let us see:

\omega =  \dfrac{m_1}{M_1*m_2}

\omega =  \dfrac{200}{132*0,821}

\omega =  \dfrac{200}{108,372}

\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark

_________________________________
_________________________________


<span>Another way to find the answer:
</span>
We have the following data: 

W (molality) = ? (in molal)
n (number of mols) = ?
m1 (solute mass) = 20 % m/m = 20g/100mL → (in g to 1L) = 200 g
m2 (disolvent mass) the remaining percentage, in the case: 80 % m/m = 800 g → m2 (disolvent mass) = 0,8 Kg
M1 (Molar mass of solute) (NH4)2 SO4 
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol 


<span>Let's find the number of mols (n), let's see:

</span>n =  \dfrac{m_1}{M_1}

n = \dfrac{200}{132}

n \approx 1,5\:mol

Now, we apply all the data found to the formula of Molality, let us see:

\omega =  \dfrac{n}{m_2}

\omega =  \dfrac{1,5}{0,8}

&#10;\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark

I hope this helps. =)
7 0
3 years ago
The density of water is 1.00 g/ml at 4°c. how many water molecules are present in 2.56 ml of water at this temperature?
o-na [289]

Answer:

             8.55 × 10²² Molecules

Solution:

Step 1: Calculate the Mass of Water as;

                       Density  =  Mass ÷ Volume

Solving for Mass,

                       Mass  =  Density × Volume

Putting values,

                       Mass  =  1 g.mL⁻¹ × 2.56 mL

                       Mass  =  2.56 g

Step 2: Calculate Moles of Water as,

                       Moles  =  Mass ÷ M.Mass

Putting values,

                       Moles  =  2.56 g ÷ 18 g.mol⁻¹

                       Moles  =  0.142 mol

Step 3: Calculate Number of Molecules of Water as,

                       # of Molecules  =  Moles × 6.022 × 10²³

Putting value of mole,

                       # of Molecules  =  0.142 × 6.022 × 10²³

                       # of Molecules  =  8.55 × 10²² Molecules

5 0
3 years ago
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