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Sholpan [36]
3 years ago
9

Explain how you can obtain hydrated sodium sulphate crystals from an aqueous solution of sodium sulphate​

Chemistry
2 answers:
Mice21 [21]3 years ago
6 0

Answer:

If you left your aqueous sodium sulfite solution open on the side, atmospheric oxygen would cause the chemical change, the water would hydrate the new compound and the slow evaporation of water would give you lovely large crystals of Na2SO4 without you having to do anything!

lana66690 [7]3 years ago
6 0

Answer:

If you left your aqueous sodium sulfite solution open on the side, atmospheric oxygen would cause the chemical change, the water would hydrate the new compound and the slow evaporation of water would give you lovely large crystals of Na2SO4 without you having to do anything!

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Urgent plzz help meeee thx
MakcuM [25]

Answer:

8

Explanation:

From the question given above, the following data were obtained:

t–butyl ion = (CH₃)₃C⁺

Number of valence electron =?

The valence electron(s) talks about the combining power of an element or compound as the case may be.

Considering the t–butyl ion, (CH₃)₃C⁺ we can see that it has a charge of +1 indicating that it has given out 1 electron to attain the stable octet configuration which has a valence electrons of 8. Thus, the valence electron of t–butyl ion, (CH₃)₃C⁺ is 8

5 0
3 years ago
What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
lutik1710 [3]
The first dissociation for H2X:
                        H2X +H2O ↔ HX + H3O
initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


4 0
3 years ago
Which type of energy transformation comes from electric heater?​
ivolga24 [154]

Answer:

electric energy ---> heat energy or A.

Explanation:

the name says it all

3 0
3 years ago
Read 2 more answers
A catalyst is used in the chemical reaction to break down hydrogen peroxide into water and oxygen. At the end of the reaction, t
Natasha_Volkova [10]
At the end of the reaction, the catalyst is UNCHANGED.
:)
8 0
3 years ago
A gas has a pressure of 5.7 atm at 100.0°C. What is its pressure at20.0°C (Assume volume is unchanged)
son4ous [18]

Answer:

\large \boxed{\text{4.5 atm}}

Explanation:

The volume and amount of gas are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}

Data:

p₁ =5.7 atm; T₁ = 100.0 °C

p₂ = ?;          T₂ =  20.0 °C

Calculations:

1. Convert the temperatures to kelvins

T₁ = (100.0 + 273.15) K = 373.15

T₂ =  (20.0 + 273.15) K = 293.15

2. Calculate the new pressure

\begin{array}{rcl}\dfrac{5.7}{373.15} & = & \dfrac{p_{2}}{293.15}\\\\0.0153 & = & \dfrac{p_{2}}{293.15}\\\\0.0153\times 293.15 &=&p_{2}\\p_{2} & = & \textbf{4.5 atm}\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.5 atm}}$}

6 0
3 years ago
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