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Misha Larkins [42]
3 years ago
8

Which arrow represents water forming outside a cold water bottle left outside

Chemistry
1 answer:
marin [14]3 years ago
7 0

Answer:

The phase transition that represent the the cooling of gaseous water to form liquid outside a bottle left outside which is a process known as is represented by the arrow 2

Explanation:

The different phase transitions that occur in mater are;

1) A phase transition from the solid phase to the liquid phase which occur by heating, represented by the arrow 2 and the process is known as melting

2)  A phase transition from the liquid phase to the gaseous phase which occur by further heating the liquid, represented by the arrow 1 and the process is known as evaporation

3) A phase transition from the gaseous phase to the liquid phase which occur by cooling the hot gas, represented by the arrow 2 and is known as condensation

4) A phase transition from the liquid phase to the solid phase which occur by further cooling the liquid, represented by the arrow 5 and the process is known as freezing

5) A phase transition from the solid phase to the gaseous phase which occur by heating the cold solid, represented by the arrow 3 and the process is known as sublimation

6) A phase transition from the gaseous phase to the solid phase which occur by cooling the gas, represented by the arrow 6

Therefore, the phase transition that represent the the cooling of gaseous water to form liquid outside a bottle left outside is represented by the arrow 2

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Element A has 15 protons. Determine the number of neutrons in its isotope with mass number 33.
mario62 [17]
33 - 15 = 18
This element is P-33 
7 0
4 years ago
A solution is prepared by dissolving 0.24 mol of butanoic acid and 0.25 mol of sodium butanoate in water sufficient to yield 1.0
beks73 [17]

Answer:

Butanoic acid present in solution

Explanation:

In this case, we have a buffer solution of butanoic acid and sodium butanoate.  In other words a reaction like this:

HC₄H₇O₂ + H₂O <------> C₄H₇O₂⁻ + H₃O⁺   Ka = 1.5x10⁻⁵

The low value of Ka means that this is a weak acid. So, after this, the NaOH is added to the solution.

The NaOH is a really strong base, so we might expect that the pH of the solution increase drastically, however this do not occur.

The reason for this is because the first thing to happen in this reaction is an acid base reaction.

The NaOH react with the butanoic acid still present in solution, because is a weak acid, so in solution, this acid is not completely dissociated into it's respective ions. So the butanoic acid reacts with the NaOH and the products:

HC₄H₇O₂ + NaOH <------> Na⁺C₄H₇O₂⁻ + H₂O

So, because of this, the pH increase but not much.

6 0
3 years ago
What is an everyday example of covalent bonds and ionic bonds
ludmilkaskok [199]

Answer:

1. CARBON DIOXIDE- it is a covalent compound, which is used in soft/cold drinks and some other fluids as well , and use it in daily life.

2. HYDROGEN MONOXIDE- it is the normal or original or pure water which we drink everyday in our daily life and it is very important for our survival

Explanation:

3 0
3 years ago
Chlorine atoms tend to form ions by taking in one electron. This means the atom has
8090 [49]

Answer:

The additional electron does not affect the atomic number nor the mass number. The atomic mass is the sum of neutrons and protons. Electrons are not accounted in atomic mass as their mass is so small, approximately 1/1,836 of a protons mass. Atomic number is the total number of protons of an atom. Chlorine taking in another electron will not change its atomic number. In an ion, the atomic number and atomic mass do not change from the original.

8 0
3 years ago
A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Assume no change in volume of t
Verdich [7]

Pb(NO₃)₂ ⇒limiting reactant

moles PbI₂ = 1.36 x 10⁻³

% yield  = 87.72%

<h3>Further explanation</h3>

Given

Reaction(unbalanced)

Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)

Required

  • moles of PbI₂
  • Limiting reactant
  • % yield

Solution

Balanced equation :

Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

mol Pb(NO₃)₂ :

= 0.45 : 331 g/mol

= 1.36 x 10⁻³

mol NaI :

= 250 ml x 0.25 M

= 0.0625

Limiting reactant (mol : coefficient)

Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³

NaI : 0.0625 : 2 = 0.03125

Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)

moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)

Mass of PbI₂ :

= mol x MW

=  1.36 x 10⁻³ x 461,01 g/mol

= 0.627 g

% yield = 0.55/0.627 x 100% = 87.72%

7 0
3 years ago
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