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mario62 [17]
3 years ago
7

When can I use ideal gas law?

Chemistry
1 answer:
Elis [28]3 years ago
3 0
Answer:

The ideal gas law can be used in stoichiometry problems in which chemical reactions involve gases. Standard temperature and pressure (STP) are a useful set of benchmark conditions to compare other properties of gases. At STP, gases have a volume of 22.4 L per mole.
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Please help with 3 and 4
mafiozo [28]
3. volume of gas because its changing because of the temperature
4. 67.6 breaths per minute
     65+73+67+71+62=338/5=67.6
                                           ^because there are five terms that we added
7 0
3 years ago
Hi there, I need help producing the aim and hypothesis for the question
svetlana [45]

Answer:

in supernova gold can be turned into lead.

its practical to obtain gold from lead

hope this helps you☺️☺️

4 0
3 years ago
The stronger the wind, the larger the particles it erodes.
Fittoniya [83]
I believe the statement above is true. The stronger the wind, the larger the particles it erodes<span>. The stronger the wind, the larger the particles that are carried away.

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7 0
3 years ago
HELP IM AB TO CRY
Lemur [1.5K]

Answer:

1.23 × 10³ N

Explanation:

Step 1: Given and required data

  • Mass of the person (m): 125 kg
  • Acceleration due to the gravitational force (g): 9.81 m/s²

Step 2: Calculate the force acting between the Earth and a 125-kg person standing on the surface of the Earth

We will use Newton's second law of motion.

F = m × g

F = 125 kg × 9.81 m/s²

F = 1.23 × 10³ N

8 0
2 years ago
Consider the reaction 2CO * O2 —&gt; 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c
Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO}  *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%

Best regards!

8 0
2 years ago
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