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3 years ago

What is the difference between a permanent change and a temporary change? What is

Chemistry

2 answers:

NikAS [45]3 years ago

6 0

Answer:

temporary changes are the changes which are there only for a short period if time.

Explanation:

generally temporary changes are reversible. permanent changes are the changes which remain for a longer time and are not reversible

Dennis_Churaev [7]3 years ago

6 0

Answer:

Permanent mangent is one that rerains its magnetic

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Balance the equation for the reaction in which sodium oxide reacts with water to form sodium hydroxide.

tensa zangetsu [6.8K]

Answer:

$➢ \: Balance \: the \: equation \: for \: the \: reaction \\ in \: which \: sodium \: oxide \: reacts \\ with \: water \: to \: form \: sodium \: hydroxide.$

⇒ We have Na2O + H2O --> NaOH. We have 2 sodiums and 2 oxygens and 2 hydrogens on the left side, but only one of each on the right side.

Sodium Oxide + Water → Sodium Hydroxide

⇒ Na2O + H2O → 2NaOH .

Sodium oxide is used in ceramics and glasses. Sodium oxide reacts exothermically with cold water to produce sodium hydroxide solution.

6 0

2 years ago

How would you prepare 100 ml of 0.4 M MgSO4 from a stock solution of 2 M MgSO4?

miss Akunina [59]

OK, so to answer this question, you will simply use the molality equation which is as follows:
<span>M1V1 = M2V2
In the givens you have:
M1 = 2M
V1 is the unknown
M2 = 0.4M
V2 = 100 ml

</span>plug in the givens in the above equation:
<span>2 x V1 = 0.4 x 100
</span>therefore:
V1 = 20 ml

Based on this: you should take 20 ml of the 2 M solution and make volume exactly 100 ml in a volumetric flask by diluting in water.

7 0

3 years ago

Calculate the molarity of a solution prepared by diluting 7.0 mL of 4.0 M solution to a volume of 30 mL.

Finger [1]

Considering the definition of dilution, the molarity of a solution prepared by diluting 7.0 mL of 4.0 M solution to a volume of 30 mL is 0.93 mL.

<h3>What is diluion</h3>

First of all, you have to know that when it is desired to prepare a less concentrated solution from a more concentrated one, it is called dilution.

Dilution is the process of reducing the concentration of solute in solution, which is accomplished by simply adding more solvent to the solution at the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

where

Ci: initial concentration
Vi: initial volume
Cf: final concentration
Vf: final volume

<h3>Molarity of the solution in this case</h3>

In this case, you know:

Ci= 4 M
Vi= 7 mL
Cf= ?
Vf= 30 mL

Replacing in the definition of dilution:

4 M× 7 mL= Cf× 30 mL

Solving:

(4 M× 7 mL)÷ 30 mL= Cf

In summary, the molarity of a solution prepared by diluting 7.0 mL of 4.0 M solution to a volume of 30 mL is 0.93 mL.

Learn more about dilution:

brainly.com/question/20113402

brainly.com/question/22762236

8 0

2 years ago

What is true of a reaction that has reached equilibrium?

BabaBlast [244]

The correct option is A.

A chemical reaction is said to have reached an equilibrium stage if the rate of reaction of the forward reaction is equal to the rate of reaction of the reverse reaction. Two way arrows are usually used to depict equilibrium reactions. These arrows indicate that the chemical reaction can move both ways. At the equilibrium point the concentrations of both the reactants and the products are equal.

4 0

3 years ago

Read 2 more answers

The solubility of carbon dioxide in water is 0.161 g CO2in 100 mL of water at 20ºC and 1.00 atmCO2. A soft drink is carbonated w

Sedbober [7]

<u>Answer:</u> The solubility of carbon dioxide at 5.50 atm is $0.886g/100mL$

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{A}=K_H\times p_{A}$

Or,

$\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}$

where,

$C_1\text{ and }p_1$ are the initial concentration and partial pressure of carbon dioxide

$C_2\text{ and }p_2$ are the final concentration and partial pressure of carbon dioxide

We are given:

$C_1=0.161g/100mL\\p_1=1.00atm\\C_2=?\\p_2=5.50atm$

Putting values in above equation, we get:

$\frac{0.161g/100mL}{C_2}=\frac{1.00atm}{5.50atm}\\\\C_2=\frac{0.161g/100mL\times 5.50atm}{1.00atm}=0.886g/100mL$

Hence, the solubility of carbon dioxide at 5.50 atm is $0.886g/100mL$

4 0

4 years ago