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Wittaler [7]
3 years ago
8

In this salt water floating egg experiment, how will you use these calculations to find the density of the egg? I’m really confu

sed and need help please!

Chemistry
1 answer:
Agata [3.3K]3 years ago
5 0

Glass 1 Freshwater  

The egg went directly to the base of the glass. Consequently it has sunk and did not drift  

237.0  

237 (1 glass)  

1.000

Glass 2  

Water with 2 teaspoons of salt  

A similar outcome for glass 2 as container 1 when egg was set into the container it sank to the base.  

248.4  

237  

1.048

Glass 3  

Water with 3 teaspoons of salt  

In reference to glass 1 and 2 the egg sank to the base for a third time.  

254.1  

237  

1.072  

Glass 4  

Water with 4 teaspoons of salt  

The egg remained gliding to where a little bit of the egg was standing out on the top, and when pushed down the egg returned up.  

259.8  

237  

1.096

Hope this helps!

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you know just in order to bring that nostalgia into chemists minds

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Based on the electron configuration of the two atoms, predict the ratio of metal cationic (+) atom to nonmetal anionic (-) atom
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The metal cationic (+) atom to nonmetal anionic (-) atom ratio in the compound formed between Potassium and Chlorine is 1:1.

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The electronic configuration of the atoms is not shown here but the metal cationic (+) atom to nonmetal anionic (-) atom ratio in the compound formed between Potassium and Chlorine is 1:1.

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6 0
2 years ago
How many milligrams of magnesium reacts with excess HCl to produce 31.2 mL of hydrogen gas at 754 Torr and 25.0℃.
Ivanshal [37]

Answer:

There will react 30.9 milligrams of magnesium

Explanation:

Step 1: Data given

Volume of hydrogen = 31.2 mL

Pressure = 754 torr = 754/760 = 0.992 atm

Temperature = 25.0 °C = 298 Kelvin

Step 2: The balanced equation

Mg + 2HCl → MgCl2 + H2

Step 3: Calculate moles of H2

p*V = n*R*T

⇒ with p = the pressure of H2 = 0.992 atm

⇒with V = the volume of H2 = 31.2 mL = 0.0312 L

⇒ with n = the moles of H2 = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T= the temperature = 25.0 °C = 298 Kelvin

n = (p*V)/(R*T)

n = (0.992*0.0312)/(0.08206*298)

n = 0.00127 moles

Step 4: Calculate moles of Mg

For 1 mol of Mg we need 2 moles of HCl to produce 1 mol of MgCl2 and 1 mol of H2

For 0.00127 moles of H2 we need 0.00127 moles of Mg

Step 5: Calculate mass of Mg

Mass of Mg = moles of Mg * molar mass of Mg

Mass of Mg = 0.00127 moles * 24.3 g/mol

Mass of Mg = 0.0309 grams = 30.9 mg of Mg

There will react 30.9 milligrams of magnesium

3 0
3 years ago
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