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lawyer [7]
3 years ago
14

Alana drew diagrams to show how particles in a spoon move before and after the spoon is placed in hot liquid.

Chemistry
2 answers:
jonny [76]3 years ago
7 0

The correct answer is:

D. Switch the labels on the diagrams.

Switching the labels on the diagrams will correct Alana’s error.

|Huntrw6|

Wewaii [24]3 years ago
7 0

Answer: Option (D) is the correct answer.

Explanation:

Before placing the spoon in a hot liquid, particles of spoon were closer to each other and vibrating at their respective positions.

But when the spoon is placed in a hot liquid then due to increase in temperature of spoon its particles gain kinetic energy. Therefore, they colloid more frequently.

Thus, we can conclude that switch the labels on the diagrams best explains how to correct Alana’s error.

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The mass is 5kg
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2 years ago
A balloon contains 269.7 L of helium at 6.12ºC and 1.00 atm. What is the temperature (in ºC) of the gas if the volume has increa
MatroZZZ [7]

Answer:

T2= 7.3°C

Explanation:

To solve this problem we will use Charles law equation i.e,

V1/T1 = V2/T2    

Given data

V1 = 269.7 L

T1 = 6.12 °C

V2= 320.4 L

T2=?

Solution:

Now we will put the values in equation

269.7 L / 6.12°C  = 320.4 L / T2

T2=  320.4 L × 6.12°C/ 269.7 L

T2= 1960.85 °C. L /269.7 L

T2= 7.3°C

3 0
3 years ago
5. Why are lonic compounds good conductors of electricity?
ziro4ka [17]

Answer:

The options are unclear, however, the correct option is:

Aqueous solutions of ionic compounds cause to dissociate, hence, ions are free to conduct electricity

Explanation:

Ionic compounds are compounds formed from ions (charged atoms). For example, NaCl is an ionic compound from the following ions; Na+ (cation) and Cl- (anion). One characteristics of ionic compounds is their ability to dissociate into the ions that form them when in an aqueous solution i.e. NaCl will dissociate into Na+ and Cl- when in an aqueous solution.

These disssociated ions are free to conduct electricity, hence, making ionic compounds good conductors of electricity.

6 0
2 years ago
Which of the following statements are TRUE for BOTH oxygen and sulfur atoms?
irina1246 [14]

Answer:

b. have the same kind number of complete shells

4 0
2 years ago
Iron (III) oxide and hydrogen react to form iron and water, like this: Fe 03(s)+3H9)2Fe(s)+3HO) At a certain temperature, a chem
belka [17]

The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound             Amount

  Fe₂O₃                     3.95 g

     H₂                        4.77 g

     Fe                        4.38 g

    H₂O                      2.00 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

<u>Answer:</u> The value of equilibrium constant for given equation is 1.0\times 10^{-4}

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M

  • <u>For water:</u>

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M

For the given chemical equation:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

The expression of equilibrium constant for above equation follows:

K_{eq}=\frac{[H_2O]^3}{[H_2]^3}

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}

Hence, the value of equilibrium constant for given equation is 1.0\times 10^{-4}

6 0
3 years ago
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