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nataly862011 [7]
2 years ago
12

A sample of 8.00 g of Mg(OH)2 is added to 24.2 mL of 0.205 M HNO3. How many moles of Mg(OH)2 our present after the reaction is c

omplete?
Chemistry
1 answer:
kogti [31]2 years ago
6 0

The amount of Mg(OH)2  present after the reaction is complete is 0.136 moles of Mg(OH)2.

The equation of the reaction is;

2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)

Number of moles of Mg(OH)2 = 8.00 g/58 g/mol = 0.138 moles

Number of moles of HNO3 = 0.205 M × 24.2 mL/1000 = 0.00496 moles

Given that;

2 moles of HNO3 reacts with 1 mole of Mg(OH)2

0.00496 moles of HNO3 reacts with 0.00496 moles ×  1 mole /2 moles = 0.00248 moles of Mg(OH)2

Hence, Mg(OH)2 is the reactant in excess.

The amount of Mg(OH)2 remaining = Amount present - Amount reacted

Hence; 0.138 moles - 0.00248 moles = 0.136 moles of Mg(OH)2

Learn more: brainly.com/question/9743981

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Answer:

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A gas has a pressure of 5.7 atm at 100.0°C. What is its pressure at20.0°C (Assume volume is unchanged)
son4ous [18]

Answer:

\large \boxed{\text{4.5 atm}}

Explanation:

The volume and amount of gas are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}

Data:

p₁ =5.7 atm; T₁ = 100.0 °C

p₂ = ?;          T₂ =  20.0 °C

Calculations:

1. Convert the temperatures to kelvins

T₁ = (100.0 + 273.15) K = 373.15

T₂ =  (20.0 + 273.15) K = 293.15

2. Calculate the new pressure

\begin{array}{rcl}\dfrac{5.7}{373.15} & = & \dfrac{p_{2}}{293.15}\\\\0.0153 & = & \dfrac{p_{2}}{293.15}\\\\0.0153\times 293.15 &=&p_{2}\\p_{2} & = & \textbf{4.5 atm}\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.5 atm}}$}

6 0
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The chemical bonding in sodium phosphate, Na3PO4, is classified as:
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<span>Ionic bonding between sodium and phosphate ions.</span>
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What is the empirical formula for a compound if a sample contains 3.72 g of P and 21.28 g of Cl
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n(Cl)=m(Cl)/M(Cl)
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Answer:

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Explanation:

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Step 2: Calculate the mass of iodine after 8.52 days

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