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valina [46]
2 years ago
13

A solution composed of 30.6g NH3 in 81.3g of H20. Calculate the mole fraction for NH3 and H20

Chemistry
2 answers:
HACTEHA [7]2 years ago
8 0

Answer: NH3 = 9/5    H20= 9/2

Explanation:

Mr of NH3 = 17g/mol

Mr of H20 = 18g/mol

NH3 = 30.6g divide by 17g/mol = 1.8 mol

                                                      = 9/5

H20 = 81.3g divide by 18g/mol = 4.5 mol  

                                                     = 9/2

irina [24]2 years ago
4 0

Answer:

0.286 moles

Explanation:

i hope this is helpful for you

You might be interested in
Ca(OH)2 (s) precipitates when a 1.0 g sample of CaC2(s) is added to 1.0 L of distilled water at room temperature. If a 0.064 g s
borishaifa [10]

Answer:

Ca(OH)2 will not precipitate because Q<Ksp

Explanation:

Ksp for Ca(OH)2 has already been stated in the question as 8.0 x 10-8mol2dm-6

The value of the reaction quotient depends heavily on the concentration of the reactants. As the initial concentration of the calcium carbide decreases considerably, the reaction quotient decreases until Q<Ksp hence the Ca(OH)2 will not precipitate from solution.

The reaction equation is:

CaC₂(s) + H₂O ⇒ Ca(OH)₂ + C₂H₂

From

Ca(OH)2= Ca2+ + 2OH-

Concentration of solution= 0.064×1/64= 1×10-3

Since [Ca2+] = 1×10-3

[OH-]= (2×10-3)^2= 4×10^-6

Hence Q= 4×10^-9

This is less than the Ksp hence the answer.

4 0
3 years ago
calculate how many milliliters of 0.142 M NaOH are needed to completely neutralize 21.4 mL of 0.294 M H2C4H4O6.
Flura [38]

Answer:

The answer to your question is 88.7 ml

Explanation:

Data

Volume = ?

Concentration of NaOH = 0.142 M

Volume of H₂C₄H₄O₆ = 21.4 ml

Concentration of H₂C₄H₄O₆ = 0.294 M

Balanced chemical reaction

               2 NaOH + H₂C₄H₄O₆  ⇒  Na₂C₄H₄O₆  +  2H₂O

1.- Calculate the moles of H₂C₄H₄O₆

Molarity = moles/volume

Solve for moles

moles = Molarity x volume

Substitution

moles = 0.294 x 21.4/1000

Result

moles = 0.0063

2.- Use proportions to calculate the moles of NaOH

              2 moles of NaOH ------------------ 1 moles of H₂C₄H₄O₆

               x                           ------------------ 0.0063 moles

               x = (0.0063 x 2) / 1

               x = 0.0126 moles of NaOH

3.- Calculate the volume  of NaOH

Molarity = moles / volume

Solve for volume

Volume = moles/Molarity

Substitution

Volume = 0.0126/0.142

Result

Volume = 0.088 L or 88.7 ml

3 0
3 years ago
22. Which of the following molecules or ions contain polar bonds?
goldfiish [28.3K]

Answer:

A 03

Explanation:

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3 0
2 years ago
What is the tempature -34°C expressed in kelvins? ​
8090 [49]

Answer:

239.15 is the answer

Explanation:

7 0
3 years ago
Read 2 more answers
Consider the reaction
SOVA2 [1]

Answer :

(a) The average rate will be:

\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s

(b) The average rate will be:

\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)

The expression for rate of reaction :

\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}

\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}

\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}

\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}

\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}

Thus, the rate of reaction will be:

\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}

<u>Part (a) :</u>

<u>Given:</u>

\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s

As,  

-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}

and,

\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}

\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s

\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s

<u>Part (b) :</u>

<u>Given:</u>

\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s

As,  

-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}

and,

-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}

\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s

\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s

5 0
3 years ago
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