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svet-max [94.6K]
3 years ago
8

At 25 celsius, 1.00 ,ole of O2 was found to occupy a volume of 12.5 L at a pressure of 198 kPa. What value of the gas constant i

n L times kPa divided by mol Times K
Chemistry
1 answer:
mr Goodwill [35]3 years ago
4 0

Answer:

R=8.301\frac{L*kPa}{mol*K}

Explanation:

Hello,

In this case, assuming oxygen as an ideal gas, it is described by:

PV=nRT

Whereas the gas constant R is required, therefore, it is computed as shown below:

R=\frac{PV}{nT} =\frac{198kPa*12.5L}{1.00mol*(25+273.15)K} \\\\R=8.301\frac{L*kPa}{mol*K}

Best regards.

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CaBr + KOH – Ca(OH), + KBr (balance first) What mass, in grams, of
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Answer:

129.73 g of CaBr₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

CaBr₂ + 2KOH –> Ca(OH)₂ + 2KBr

Next, we shall determine the mass of CaBr₂ that reacted and the mass of Ca(OH)₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of CaBr₂ = 40 + (80×2)

= 40 + 160

= 200 g/mol

Mass of CaBr₂ from the balanced equation = 1 × 200 = 200 g

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= 74 g/mol

Mass of Ca(OH)₂ from the balanced equation = 1 × 74 = 74 g

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From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Finally, we shall determine the mass of CaBr₂ that react when 48 g of Ca(OH)₂ were produced. This can be obtained as follow:

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Therefore, Xg of CaBr₂ will react to produce 48 g of Ca(OH)₂ i.e

Xg of CaBr₂ = (200 × 48)/74

Xg of CaBr₂ = 129.73 g

Thus, 129.73 g of CaBr₂ were consumed.

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