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Kaylis [27]
3 years ago
15

An arctic weather balloon is filled with 38.5 L of helium gas inside a prep shed. The temperature inside the shed is 8. °C. The

balloon is then taken outside, where the temperature is -41. °C. Calculate the new volume of the balloon. You may assume the pressure on the balloon stays constant at exactly 1 atm. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
Naily [24]3 years ago
3 0

Answer: The new volume of the balloon is 197.31 L.

Explanation:

Given: V_{1} = 38.5 L,        T_{1} = 8^{o}C

V_{2} = ?,                     T_{2} = 41^{o}C

According to Charles law, at constant pressure the volume of a gas is directly proportional to temperature.

Formula used to calculate the new volume is as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{38.5 L}{8^{o}C} = \frac{V_{2}}{41^{o}C}\\V_{2} = \frac{38.5 L \times 41^{o}C}{8^{o}C}\\= 197.31 L

Thus, we can conclude that the new volume of the balloon is 197.31 L.

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Anvisha [2.4K]

Answer:

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7 0
2 years ago
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Magnesium (average atomic mass = 24.3050 amu) consists of three isotopes with masses 23.9850 amu, 24.9858 amu, and 25.9826 amu.
iris [78.8K]

Answer: The first isotope has a relative abundance of 79% and last isotope has a relative abundance of 11%

Explanation: Given that the average atomic mass(M) of magnesium

= 24.3050amu

Mass of first isotope (M1) = 23.9850amu

Mass of middle isotope (M2)=24.9858amu

Mass of last isotope(M3)= 25.9826amu

Total abundance = 1

Abundance of middle isotope = 0.10

Let abundance of first and last isotope be x and y respectively.

x+0.10+y =1

x = 0.90-y

M = M1 × % abundance of first isotope + M2 × % of middle isotope +M3 ×% of last isotope

24.03050= 23.985× x + 24.9858 ×0.10 + 25.9826×y

Substitute x= 0.90-y

Then

y = 0.11

Since y=0.11, then

x= 0.90-0.11

x=0.79

Therefore the relative abundance of the first isotope = 11% and the relative abundance of the last isotope = 79%

5 0
3 years ago
What element is in group 13, period 4
Ludmilka [50]

Answer:

Gallium

Explanation:

6 0
3 years ago
Read 2 more answers
For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi
natka813 [3]

<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ....(1)

  • <u>For Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol

For the given chemical reaction:

2Fe(s)+O_2(g)\rightarrow 2FeO(s)

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = \frac{2}{2}\times 0.0771=0.0771mol of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:  

0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g

To calculate the percentage yield of iron (ii) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0
3 years ago
Need help with this ASAP<br><br> Thanks!
olga_2 [115]

Answer:

Decomposers (either Secondary Consumer or Tertiary Consumer)

Explanation:

Decomposers eat dead materials and break them down into chemical parts. ... They keep the ecosystem free of the bodies of dead animals or carrion. They break down the organic material and recycle it into the ecosystem as nutrients. Vultures, Blowflies, hyenas, crabs, lobsters and eels are examples of scavengers.

6 0
3 years ago
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