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kompoz [17]
2 years ago
9

You have 3.50 l of solution that contains 90.0 g of sodium chloride, nacl. what is the molar mass of nacl? what is the molarity

of that solution?
Chemistry
1 answer:
nadya68 [22]2 years ago
3 0
M(NaCl) = Ar(Na) + Ar(Cl) · g/mol.
M(NaCl) = 22,99 + 35,45 · g/mol.
M(NaCl) = 58,44 g/mol.
V(NaCl) = 3,5 L.
m(NaCl) = 90 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 90 g ÷ 58,44 g/mol.
n(NaCl) = 1,54 mol.
c(NaCl) = n(NaCl) ÷ V(NaCl).
c(NaCl) = 1,54 mol ÷ 3,5 L.
c(NaCl) = 0,44 mol/l = 0,44 M.
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Elanso [62]

Answer:Material

Coefficient

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chromium

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titanium

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Drag the materials into the correct order from the material that will slide down the ramp the fastest to the material that will slide down the ramp the

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chromium

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aluminum

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Explanation:

7 0
2 years ago
120 grams of calcium nitrite Ca(NO2)2 is dissolved in a 240 mL solution. What is the molarity of the solution???
expeople1 [14]

Hey there!:

Molar mass Ca(NO2)2 = 132.089 g/mol

Mass of solute = 120 g

Number of moles:

n = mass of solute / molar mass

n = 120 / 132.089

n = 0.0009084 moles of Ca(NO2)2

Volume in liters of solution :

240 mL / 1000 => 0.24 L

Therefore:

Molarity = number of moles / volume of solution

Molarity = 0.0009084 / 0.24

Molarity = 0.003785 M

Hope that helps!

5 0
2 years ago
Consider the reaction. X ( g ) + Y ( g ) − ⇀ ↽ − Z ( g ) K p = 1.00 at 300 K In which direction will the net reaction proceed fo
marta [7]

Answer:

Explanation:

We have in this question the equilibrium

X ( g ) + Y ( g ) ⇆  Z ( g )

With the equilibrium contant Kp = pZ/(pX x pY)

The moment we change the concentration of Y, we are changing effectively the partial pressure of Y since pressure and concentration are directly proportional

pV = nRT ⇒ p = nRT/V and n/V is molarity.

Therefore we can calculate the reaction quotient Q

Qp = pZ/(pX x pY) = 1/ 1  x 0.5 atm = 2

Since Qp is greater than Kp the system proceeds from right to left.

We could also arrive to the same conclusion by applying LeChatelier´s principle which states that any disturbance in the equilibrium, the system will react in such a way to counteract the change to restore the equilibrium. Therefore, by having reduced the pressure of Y the system will react favoring the reactants side increasing some of the y pressure until restoring the equilibrium Kp = 1.

4 0
2 years ago
Two students are working together to build two models. Both models will represent the molecular structure of sodium bicarbonate,
vfiekz [6]

Answer:

Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.

<em>Note: Since no specific color was stated for oxygen atoms, the answer assigns blue colored jellybeans to represent oxygen atoms.J</em>

Explanation:

Sodium bicarbonate, NaHCO₃ is a compound composed of one atom of sodium, one atom of hydrogen, one atom of carbon and three atoms of oxygen.

Since red jellybeans represent sodium atoms, white jellybeans represent hydrogen atoms, black jellybeans represent carbon atoms and blue jellybeans represent oxygen atoms, each of the two students will require the following number of each jellybean for their model of sodium carbonate: One red jellybean, one white jellybean, one black jellybean and three blue jellybeans.

Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.

8 0
2 years ago
A 0.15 m solution of chloroacetic acid has a ph of 1.86. What is the value of ka for this acid?
dem82 [27]

Answer: 1.67\times 10^{-3}

Explanation:

ClCH_2COOH\rightarrow ClCH_2COO^-+H^+

   cM              0             0

c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Given:  c = 0.15 M

pH = 1.86

K_a = ?

Putting in the values we get:

Also pH=-log[H^+]

1.86=-log[H^+]

[H^+]=0.01

[H^+]=c\times \alpha

0.01=0.15\times \alpha

\alpha=0.06

As [H^+]=[ClCH_2COO^-]=0.01

K_a=\frac{(0.01)^2}{(0.15-0.15\times 0.06)}

K_a=1.67\times 10^{-3]

Thus the vale of K_a for the acid is 1.67\times 10^{-3}

4 0
2 years ago
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