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ikadub [295]
3 years ago
11

If 5.00 g of O2 gas has a volume of 7.20 L at a certain temperature and pressure, what volume does 15.0 g of O2 have under the s

ame conditions?
Chemistry
1 answer:
tigry1 [53]3 years ago
4 0
65.0 L.








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How many atoms of magnesium metal are needed to produce 3.87 moles of magnesium oxide
Ket [755]

Answer:

48.42g of Mg

Explanation:

4 0
2 years ago
Read 2 more answers
Can someone help Plz
Marrrta [24]

Answer: A. 1.60 liters

Explanation:

2Na_2O_2+2CO_2\rightarrow 2Na_2CO_3+O_2

As can be seen from the balanced chemical equation, 2 moles of CO_2 produce 1 mole of O_2.

According to Avogadro's law, every 1 mole of the gas occupies 22.4 liters at STP.

Thus 2 moles of CO_2 occupies 22.4\times 2=44.8L at STP and produce 1 mole of O_2 i.e. 22.4 L

44.8 L of CO_2 will produce 22.4 L of O_2

Thus 3.20L of CO_2 will produce \frac{22.4}{44.8}\times 3.20=1.60L



6 0
2 years ago
What is the product if an atom of Po-209<br> undergoes alpha decay?
g100num [7]
<h3>Answer:</h3>

Lead-205 (Pb-205)

<h3>Explanation:</h3>

<u>We are given;</u>

  • An atom of Po-209

We are supposed to identify its product after an alpha decay;

  • Polonium-209 has a mass number of 209 and an atomic number of 84.
  • When an element undergoes an alpha decay, the mass number decreases by 4 while the atomic number decreases by 2.
  • Therefore, when Po-209 undergoes alpha decay it results to the formation of a product with a mass number of 205 and atomic number of 82.
  • The product from this decay is Pb-205, because Pb-205 has a mass number of 205 and atomic number 82.
  • The equation for the decay is;

²⁰⁹₈₄Po → ²⁰⁵₈₂Pb + ⁴₂He

  • Note; An alpha particle is represented by a helium nucleus, ⁴₂He.

7 0
3 years ago
An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.
Tju [1.3M]

Answer:

a) \Delta S

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

\delta S=\frac{\delta Q}{T}

\Delta S=\frac{Q}{T}

The energy balance:

\delta U=[tex]\delta Q- \delta W

If the process is isothermical the U doesn't change:

0=[tex]\delta Q- \delta W

\delta Q= \delta W

Q= W

The work:

W=\int_{V1}^{V2}P*dV

If it is an ideal gas:

P=\frac{n*R*T}{V}

W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV

Solving:

W=n*R*T*ln(V2/V1)

Replacing:

\Delta S=\frac{n*R*T*ln(V2/V1)}{T}

\Delta S=n*R*ln(V2/V1)}

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

\Delta S

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

7 0
3 years ago
Which source would be the most reliable to validate a claim?
Mekhanik [1.2K]

Answer:

In general, print publications with authors and listed sources tend to be reliable because they provide sources which readers can verify. Likewise, Web postings with a .gov (posted by the United States government) are both current and reliable.

Explanation:

4 0
3 years ago
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