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DedPeter [7]
3 years ago
12

Boron trifluoride gas is collected at in an evacuated flask with a measured volume of . When all the gas has been collected, the

pressure in the flask is measured to be . Calculate the mass and number of moles of boron trifluoride gas that were collected. Round your answer to significant digits.
Chemistry
1 answer:
Vaselesa [24]3 years ago
8 0

Answer:

0.683 mol

46.3 g

Explanation:

There is some info missing. I think this is the original question.

<em>Boron trifluoride gas is collected at 21.0 °C in an evacuated flask with a measured volume of 50.0 L. When all the gas has been collected, the pressure in the flask is measured to be 0.330 atm. Calculate the mass and number of moles of boron trifluoride gas that were collected. Round your answer to 3 significant digits.</em>

<em />

Step 1: Convert the temperature to the Kelvin scale

We will use the following expression.

K = °C + 273.15

K = 21.0°C + 273.15

K = 294.2 K

Step 2: Calculate the moles of boron trifluoride gas

We will use the ideal gas equation.

P \times V = n \times R \times T\\n = \frac{P \times V}{R \times T}  = \frac{0.330atm \times 50.0L}{\frac{0.0821atm.L}{mol.K}  \times 294.2K} = 0.683 mol

Step 3: Calculate the mass of boron trifluoride gas

The molar mass of BF₃ is 67.81 g/mol.

0.683 mol \times \frac{67.81g}{mol} = 46.3 g

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Na2CO3<span> + 2HCl === 2NaCl + H2O + CO2
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Read 2 more answers
A certain ionic compound X has a solubility in water of 40.3 g/L at 20. degrees C. Calculate the mass X of required to prepare 5
tino4ka555 [31]

Answer:

20.1 g

Explanation:

The solubility indicates how much of the solute the solvent can dissolve. A solution is saturated when the solvent dissolved the maximum that it can do, so, if more solute is added, it will precipitate. The solubility varies with the temperature. Generally, it increases when the temperature increases.

So, if the solubility is 40.3 g/L, and the volume is 500 mL = 0.5 L, the mass of the solute is:

40.3 g/L = m/V

40.3 g/L = m/0.5L

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7 0
3 years ago
PLEASE HELP!! I REALLY NEED HELP
Allushta [10]

Answer:

Explanation:

1. find the molar mass (amu) of each element and add them to get the whole molar mass.

2. divide the 1 element molar mass with the whole molar mass

3. multiple by 100 and that gives you the % composition.

<h2><u><em>56-57: NaCl</em></u></h2>

1. Na(22.99amu) + Cl (35.453amu)=58.443

2(Na):   \frac{22.99}{58.443} = .393

2(Cl): \frac{35.453}{58.443}= .607

3(Na): .393 * 100=39.3%

3(Cl): .607 * 100= 60.7%

<h2><u>58-60 </u>K_{2} CO_{3}<u /></h2>

1. K: (39.098)(2)=78.196

_ C: (12.011)(1)= 12.011

_O: (15.99)(3) = 47.997

78.196+12.011+47.997= 138.204

2:K: \frac{78.196}{138.204}= .566 <u>Step </u>3: (.566)(100)= 56.6%

2: C: \frac{12.011}{138.204}= .087 <u>Step 3</u>: (.087)(100)= 8.7%

2: O: \frac{47.997}{138.204}= .347 <u>Step 3</u>: (.347)(100) = 34.7%

<h2>61-62 Fe_{3} O_{4}</h2>

1. Fe (55.845)(3)= 167.535

_ O (15.999)(4) = 63.996

167.535+63.996=231.531

2: Fe: \frac{167.535}{231.531}= .724 Step 3: (.724)(100)= 72.4%

2: O : \frac{63.996}{231.531}= .276 Step 3: (.276)(100) = 27.6%

<h2>63-65 C_{3}H_{5}(OH)_{3}</h2>

1.

C(12.011*3)=36.033

H(1.008*5)=5.04 + (1.008*3)=3.024 so its 8.064

O(15.999*3)=47.997

add them: 92.094

2: C: \frac{36.033}{92.094}= .391 Step 3: (.391)(100) = 39.1%

2: H: \frac{8.064}{92.094}= .088 step 3: (.088)(100) = 8.8%

2: O: \frac{47.997}{92.094} = .521 step 3: (.521)(100) = 52.1%

3 0
3 years ago
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