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Radda [10]
2 years ago
4

Soda and other carbonated beverages seem more "fizzy" when they are served on an airplane at standard cruising altitude. Write a

hypothesis that could explain this observation
Chemistry
2 answers:
AveGali [126]2 years ago
7 0

Answer:

If the pressure in the air around a carbonated beverage decreases, then the beverage will seem more fizzy because more gas can come out of the liquid

Explanation:

Mekhanik [1.2K]2 years ago
3 0
Carbonated beverages have CO2 dissolved in them under pressure. As this pressure is released or reduced, then this gas can dissipate and thus cause fizziness as it escapes from the drink. At a standard cruising altitude, which would presumably be quite high, the air pressure is less than at sea level so the gas can escape easier than at a lower elevation where the air pressure is higher.
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Describe how you would prepare approximately 2 l of 0.050 0 m boric acid, b(oh)3.
Elenna [48]

The given concentration of boric acid = 0.0500 M

Required volume of the solution = 2 L

Molarity is the moles of solute present per liter solution. So 0.0500 M boric acid has 0.0500 mol boric acid present in 1 L solution.

Calculating the moles of 0.0500 M boric acid present in 2 L solution:

2 L * \frac{0.0500 mol B(OH)_{3} }{1 L} = 0.100 mol B(OH)_{3}

Converting moles of boric acid to mass:

0.100 mol B(OH)_{3} * \frac{61.83 g}{mol B(OH)_{3}}   = 6.183 g

Therefore, 6.183 g boric acid when dissolved and made up to 2 L with distilled water gives 0.0500 M solution.


5 0
2 years ago
A gas occupies 22.4 l at stp and 14.5 l at 100c and 2.00 atm pressure. how many moles of gas did the system gain or lose?
azamat
<span>At standard temperature and pressure 22.4 l of an ideal gas would contain 1 mole. in order to find the change in moles we must look at the ideal gas law PV=nRT where P=Pressure V=volume n=Moles R= Gas constant T= Temperature. To simplify this equation we will be using the gas constant at .08206 L-atm/mol-K. We must first convert 100c to k which is 373.15. Then we can plug the values into our equation which gives us (2atm)(14.5 l)=(n)(.08206 L-atm/mol-K)(373.15). After some basic algebra we get the moles to equal roughly .95 which is .05 moles less than our original system.</span>
6 0
2 years ago
A chemist heats 100.0 g of FeSO4 x 7H2O in a crucible to drive off the water. If all the water is driven off, what is the mass o
Ierofanga [76]
FeSO₄*7H₂O(s) = FeSO₄(s) + 7H₂O(g)

M(FeSO₄*7H₂O)=278.0 g/mol
M(FeSO₄)=151.9 g/mol

m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)=m(FeSO₄)/M(FeSO₄)

m(FeSO₄)=M(FeSO₄)m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)

m(FeSO₄)=151.9*100.0/278.0=54.6 g

m(FeSO₄)=54.6 g


4 0
2 years ago
Read 2 more answers
Which of the following shows an accurate combustion reaction?
Jet001 [13]

Answer:

  • <em>Cu + O₂  → CuO₂</em>

Explanation:

A <em>combustion reaction</em> is the reaction with oxygen along with the release of energy in form of heat or light.

Organic compounds (like CH₄) undergo combustion forming water and CO₂.

The combustion reaction of CH₄ is:

  • CH₄ + 2O₂ → CO₂ + 2H₂O

Hence, the first equation from the choices is not showing the combustion reaction of CH₄.

Not only organic compounds can undergo combustion. Metals and no metals can undergo combustion, i.e. metals and no metals can react with oxygen releasing light or heat.

The reaction of copper and oxygen (second choice) is a combustion reaction:

  • <em>Cu + O₂ → CuO₂</em>

The formation of water (2H₂ + O₂ → 2H₂O) is other example of a combustion reaction where no organic compounds are involved.

On the other hand, the other two equations from the choice list are not reactions with oxygen, so they do not show combustion reactions.

5 0
2 years ago
Mr. Rutherford's chemistry class was collecting data in a neutralization study. Each group had 24 test tubes to check each day f
ycow [4]
I’m pretty sure it is A at least that’s what we did at our school to test this
6 0
2 years ago
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