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OleMash [197]
2 years ago
11

450.0 mL of an unknown gas has its pressure decreased from 0.07 atm to 0.04 atm. Assuming temperature remains constant, what wou

ld be the final volume of the unknown gas
Chemistry
1 answer:
tino4ka555 [31]2 years ago
4 0

The final volume of the unknown gas is 787.5 ml

Given:

volume of unknown gas = 450.0 mL

initial pressure of unknown gas = 0.07 atm

final pressure of unknown gas = 0.04 atm

To Find:

final volume of the unknown gas

Solution:

Substituting these values into Boyle’s law, we get

P1V1 = P2V2

(0.07)(450) = (0.04)V2

V2 = (0.07)(450)/(0.04)

V2 = 787.5

So, final volume of the unknown gas is 787.5 ml

Learn more about Volume here:

brainly.com/question/25736513

#SPJ4

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san4es73 [151]

Answer:

You cannot make observations if you are 57 seconds late into the lab.

Explanation:

The atomic nucleus can split by decay into 2 or more particles as a result of the instability of its atomic nucleus due to the fact that radioactive elements possess an unstable atomic nucleus.

Now, the primary particles which are emitted by radioactive elements in order to make them decay are alpha, beta & gamma particles.

The half life equation is;

N_t = N₀(½)^(t/t_½)

Where:

t = duration of decay

t_½ = half-life

N₀ = number of radioactive atoms initially

N_t = number of radioactive atoms remaining after decay over time t

We are given;

t = 57 secs

N₀ = 100 g

Now, half life of Nitrogen-16 from online sources is 7.2 seconds. t_½ = 7.2

Thus;

N_t = 100(1/2)^(57/7.2)

N_t = 0.4139g

We are told that In order to make observations, you require at least .5g of material.

The value of N_t you got is less than 0.5g, therefore you cannot make observations if you are 57 seconds late.

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3 years ago
Al2(SO4)3 + CaCl2 → AlCl3 + CaSO4
wolverine [178]
Al2(SO4)3 + 3CaCl2 > 2AlCl3 + 3CaSO4
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compared to the energy released per mole of reactant during chemical reactions the energy released per mole of reactant during n
Zolol [24]

Answer:

\boxed{\text{Roughly a million times greater}}.

Explanation:

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The atomic orbitals of the central atom mix to form hybrid orbitals are one s and three p.

<h3>Which atomic orbitals are used to form hybrid orbitals?</h3>

Pauling supposed that in the presence of four hydrogen atoms, the s and p orbitals form four equivalent combinations which he called hybrid orbitals.

<h3>How many bonds does PF5?</h3>

Phosphorus pentafluoride has 5 regions of electron density around the central phosphorus atom (5 bonds, no lone pairs).

The resulting shape is a trigonal bipyramidal in which three fluorine atoms occupy equatorial and two occupy axial positions.

Learn more about PF5 here:

<h3>brainly.com/question/1565926</h3><h3 /><h3>#SPJ4</h3>
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Answer:

full

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