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mixas84 [53]
2 years ago
10

What mass of solid NaOH (97.0 % by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 1

0.0% solution is 1.109 g/ml.
Chemistry
1 answer:
Lelechka [254]2 years ago
3 0

<u>Answer:</u> The mass of 97 % of NaOH solution required is 114.33 g

<u>Explanation:</u>

To calculate mass of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

We are given:

Density of 10 % solution = 1.109 g/mL

Volume of 10% solution = 1 L = 1000 mL     (Conversion factor:  1 L = 1000 mL)

Putting values in above equation, we get:

1.109g/mL=\frac{\text{Mass of }10\%\text{ solution}}{1000mL}\\\\\text{Mass of }10\%\text{ solution}=1109g

The mass of 10 % solution is 1109 g.

To calculate the mass of concentrated solution, we use the equation:

c_1m_1=c_2m_2

where,

c_1\text{ and }m_1 are the concentration and mass of concentrated solution.

c_2\text{ and }m_2 are the concentration and mass of diluted solution.

We are given:

c_1=97\%\\m_1=?g\\c_2=10\%\\m_2=1109g

Putting values in above equation, we get:

97\times m_1=10\times 1109\\\\m_1=114.33g

Hence, the mass of 97 % of NaOH solution required is 114.33 g

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Answer:

4.86×10^23 molecule of Pb

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So:

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Hydrogen bonds are approximately _____% of the bond strength of covalent c-c or c-h bonds.
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Hydrogen bonds are approximately 5% of the bond strength of covalent C-C or C-H bonds.
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What volume of a 0.716 m kbr solution is needed to provide 30.5 g of kbr?
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A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600. g of water at 20.0∘C. If the heat capacity of the bomb c
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Answer:

22.7

Explanation:

First, find the energy released by the mass of the sample. The heat of combustion is the heat per mole of the fuel:

ΔHC=qrxnn

We can rearrange the equation to solve for qrxn, remembering to convert the mass of sample into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:

qrxn=−(qwater+qbomb)

The heat absorbed by the water can be calculated using the specific heat of water:

qwater=mcΔT

The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:

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Combine both equations into the first equation and substitute the known values, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute the terms of each multiplication and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)–(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)–(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)–50208 J+(420. J∘C×Tfinal)–8400 J]

Add the like terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, solve for Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

The answer should have three significant figures, so round to 22.7∘C.

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