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Gre4nikov [31]
2 years ago
12

A 50.0 mL sample if 0.0500 M Ca(OH)2 is added to 10.0 mL of 0.0200M HNO3. How many miles of excess acid or base are present?

Chemistry
1 answer:
kogti [31]2 years ago
7 0
Ca(OH)2+2HNO3=Ca(NO3)2+2H20

M =   \frac{n}{V}  \:


0.05= \frac{x}{0.05}  \\ x =0.0025 \: mol \: of \: Ca(OH)2 \:
0.02 =  \frac{y}{0.01} \\ y = 0.002 \: mol \: of \: HNO3 \:

HNO3 acid is limit
so,
0.0005 mol of Ca(OH)2 is excess
or
0.037 g of Ca(OH)2 is excess
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Answer:

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We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.

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masya89 [10]

Answer:

A

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5.1169 mol of Ne is held at 0.9148 atm and 911 K. What is the volume of its container in liters?
sveticcg [70]

By applying the Boyle's equation and substituting our given data the volume of the container was found to be 418.14 Litres

<h3>Boyle's  Law</h3>

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We know that the relationship between pressure and temperature is given as

PV = nRT

R = 0.08206

Making the volume subject of formula we have

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Substituting our given data to find the volume we have

V = 5.1169*0.08206*911/0.9148

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Learn more about Boyle's law here:

brainly.com/question/469270

5 0
1 year ago
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Lana71 [14]
<h3>Answer:</h3>

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<h3>Explanation:</h3>
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In this case, we are required to convert 0.3227 MW to kilowatts

We need to know that;

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Thus, the peak power output is 322.7 kW

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