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Fed [463]
3 years ago
11

To make 4.25 L of 3.0 M NaOH solution how much of a 9.0 M NaOH solution would be needed?

Chemistry
1 answer:
Elan Coil [88]3 years ago
4 0

Answer:

1.41 L

Explanation:

Initially the concentration of NaOH was 9.0M (say M1) finally the concentration becomes 3.0M (say M2)

intial volume (say V1) is to be calculated when the final volume (say V2) is 4.25L

we have,

M1V1=M2V2

or V1=(M2V2)/M1= 3.0×4.25/9.0= 1.41L

Hence 1.41 L of 9.0M solution is required to prepare 4.25L of 3.0M NaOH solution

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A 30.7 g sample of Strontium nitrate, Sr(NO3)2•nH2O, is heated to a constant mass of 22.9 g. Calculate the hydration number.
Elodia [21]

Answer:

  • <em>Hydration number:</em> 4

Explanation:

<u>1) Mass of water in the hydrated compound</u>

Mass of water = Mass of the hydrated sample - mass of the dehydrated compound

Mass of water = 30.7 g - 22.9 g = 7.8 g

<u>2) Number of moles of water</u>

  • Number of moles = mass in grams / molar mass

  • molar mass of H₂O = 2×1.008 g/mol + 15.999 g*mol = 18.015 g/mol

  • Number of moles of H₂O = 7.9 g / 18.015 g/mol = 0.439 mol

<u>3) Number of moles of Strontium nitrate dehydrated, Sr (NO₃)₂</u>

  • The mass of strontium nitrate dehydrated is the constant mass obtained after heating = 22.9 g

  • Molar mass of Sr (NO₃)₂ :  211.63 g/mol (you can obtain it from a internet or calculate using the atomic masses of each element from a periodic table).

  • Number of moles of Sr (NO₃)₂ = 22.9 g / 211.63 g/mol =  0.108 mol

<u>4) Ratio</u>

  • 0.439 mol H₂O / 0.108 mol Sr(NO₃)₂ ≈  4 mol H₂O : 1 mol Sr (NO₃)₂

Which means that the hydration number is 4.

4 0
4 years ago
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1. At what temperature dose water freeze
Rus_ich [418]

Answer:

1. water will freeze at a temperature below 32 degrees fahrenheit 0 degree celsius.

2. Ice will melt at a temperature above 32 degrees fahrenheit 0 degrees celsius.

3. water boils at 212 degrees fahrenheit or 100 degrees celsius.

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Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the following equation: CO2(s)→C
GREYUIT [131]

Answer:

m=8.79kg

Explanation:

First of all we need to calculate the heat that the water in the cooler is able to release:

Q=\rho * V*Cp*\Delta T

Where:

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  • V is the volume
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Q=1 g/cm^3 *15000 cm^3*4.184 \frac{J}{g*^{\circ}C}*(10-90)^{\circ}C

Q=-5020800 J=-5020.8 kJ

To calculate the mass of CO2 that sublimes:

-Q=\Delta H_{sub}*m

Knowing that the enthalpy of sublimation for the CO2 is: \Delta H_{sub}=571 kJ/kg

5020.8 kJ=571 kJ/kg*m

m=\frac{5020.8 kJ}{571 kJ/kg}=8.79kg

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Answer:

it is b

Explanation:

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How is gas different from a liquid
satela [25.4K]
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4 years ago
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