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2 years ago

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You weigh a dry flask, and place a chunk of solid into it, then weigh it again. You then fill the flask to the top with water an

d weigh the full flask. Finally, you take the solid out, fill the flask to the top with water and weigh it once more. Mass of flask in (g) 58.5591 Mass of flask plus solid (g) 134.2642 Mass of flask plus solid and water (g) 142.3225 Mass of flask filled with water (g) 83.8179 Density of water at the measured water temperature (g/cc) 0.9946 Calculate the following: The mass of the solid in (g)

1 answer:

GaryK [48]2 years ago

6 0

Note: You only asked for the mass of the solid, if you have further questions, you can ask, I will give you the solution

Answer:

Mass of solid = 75.7051 g

Explanation:

Mass of flask plus + solid = 134.2642 g

Mass of flask = 58.5591 g

Mass of solid = (Mass of flask + solid) - (Mask of flask)

Mass of solid = ( 134.2642) - (58.5591)

Mass of solid = 75.7051 g

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(4) homobegenous mixture is your answer.

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A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.020 h, and the assem

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Answer: The workdone W = 505J

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2 years ago

Suppose you are studying the K sp of K C l O 3 , which has a molar mass of 122.5 g/mol, at multiple temperatures. You dissolve 4

Ghella [55]

Answer:

The K sp Value is $K_{sp}=7.40$

Explanation:

From the question we are told that

The of $KClO_3$ is = 122.5 g/ mol

The mass of $KClO_3$ dissolved is $m_s = 4.0g$

The volume of solution is $V_s = 12mL = 12*10^{-3}L$

The number of moles of $KClO_3$ is mathematically evaluated as

$No \ of \ moles \ = \frac{mass }{Molar \ mass}$

Substituting values

$= \frac{4}{122.5}$

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$concentration = \frac{No \ of \ moles}{volume }$

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The solubility product constant is mathematically represented as

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Since there is no ionic reactant we have

$K_{sp} = [k^+] [ClO_3^-]$

$= Z^2$

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2 years ago

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What is the net ionic equation of the reaction of MgCl2 with NaOH? Express your answer as a chemical equation. View Available Hi

kari74 [83]

Answer:

Net ionic equation for the reaction between MgCl₂ and NaOH in water:

$\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s)$ .

Net ionic equation for the reaction between MgSO₄ and BaCl₂ in water:

$\rm {Ba}^{2+}\; (aq) + {SO_4}^{2-}\;(aq) \to BaSO_4\; (s)$ .

Explanation:

Start by finding the chemical equations for each reaction:

MgCl₂ reacts with NaOH to form Mg(OH)₂ and NaCl. This reaction is a double decomposition reaction (a.k.a. double replacement reaction, salt metathesis reaction.) This reaction is feasible because one of the products, Mg(OH)₂, is weakly soluble in water and exists as a solid precipitate.

$\rm MgCl_2\; (aq) + 2\; NaOH\; (aq)\to Mg(OH)_2 \; (s) + 2\; NaCl\; (aq)$ .

MgSO₄ reacts with BaCl₂ in a double decomposition reaction to produce BaSO₄ and MgCl₂. Similarly, the solid product BaSO₄ makes this reaction is feasible.

$\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq)$ .

How to rewrite a chemical equation to produce a net ionic equation?

Rewrite all reactants and products that ionizes completely in the solution as ions.
Eliminate ions that exist on both sides of the equation to produce a net ionic equation.

Typical classes of chemicals that ionize completely in water:

Soluble salts,
Strong acids, and
Strong bases.

Keep the formula of salts that are not soluble in water, weak acids, weak bases, and water unchanged.

Take the first reaction as an example, note the coefficients:

MgCl₂ is a salt and is soluble in water. Each unit of MgCl₂ can be written as $\rm Mg^{2+}$ and $\rm 2\; Cl^{-}$ .
NaOH is a strong base. Each unit of NaOH can be written as $\rm Na^{+}$ and $\rm OH^{-}$ .
Mg(OH)₂ is a weak base and should not be written.
NaCl is a salt and is soluble in water. Each unit of NaCl can be written as $\rm Na^{+}$ and $\rm Cl^{-}$ .

$\rm Mg^{2+} + 2\; Cl^{-} + 2\; Na^{+} + 2\; OH^{-} \to Mg(OH)_2\;(s) + 2\; Na^{+} + 2\; Cl^{-}$ .

Ions on both sides of the equation:

$\rm 2\; Cl^{-}$ , and
$\rm 2\; Na^{+}$ .

Add the state symbols:

$\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s)$ .

For the second reaction:

$\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq)$ .

$\rm Mg^{2+} + 2\; {SO_4}^{2-} + Ba^{2+} + 2\; Cl^{-} \to BaSO_4\; (s) + Mg^{2+} + 2\; Cl^{-}$ .

$\rm Ba^{2+}\; (aq) + {SO_4}^{2-}\; (aq) \to BaSO_4\; (s)$ .

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2 years ago

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kupik [55]

The question provides the data in an incorrect way, but what the question is asking is for the entropy change when combining 3 moles of water at 0 °C (273.15 K) with 1 mole of water at 100 °C (373.15 K). We are told the molar heat capacity is 75.3 J/Kmol. We will be using the following formula to calculate the entropy change of each portion of water:

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C = molar heat capacity
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We can first find the equilibrium temperature of the mixture which will be the value of T₂ in each case:

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Now we can find the change in entropy for the 3 moles of water heating up to 298.15 K and the 1 mole of water cooling down to 298.15 K:

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ΔS = (1 mole)(75.3 J/Kmol)ln(298.15/373.15)
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Now we combine the entropy change of each portion of water to get the total entropy change for the system:

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The entropy change for combining the two temperatures of water is 2.9 J/K.

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2 years ago

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