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1 year ago

Identify the precipitate that forms (if any) when aqueous solutions of barium nitrate and sodium sulfate are mixed.

Chemistry

1 answer:

Alborosie1 year ago

4 0

The precipitate that forms when aqueous solutions of barium nitrate and sodium sulfate are mixed is barium sulphate.

Barium Nitrate is a colorless to white, odorless, crystalline powder. It is used in making fireworks, for green lights and neon lights, and in ceramic glazes.

Sodium Sulfate Anhydrous is the anhydrous, sodium salt form of sulfuric acid. Sodium ion is the principal cation of the extracellular fluid and plays a large part in the therapy of fluid and electrolyte disturbances.

When aqueous solutions of barium nitrate and sodium sulfate are mixed is barium sulphate, one soluble salt, sodium nitrate, and an insoluble salt, barium sulphate.

Learn more about barium nitrate, click here brainly.com/question/9597126

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Scientific Notation - PLEASE HELP!

zmey [24]

Answer:

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3 years ago

A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a

Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

$H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)$

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = $m C .$ Δ $T$

m is mass of the solution = 151.8 mL * $\frac{1 g}{1 mL} = 151.8 g$

C is the specific heat of solution = 4.18 $\frac{J}{g. ^{0}C}$

ΔT is the temperature change = $31.50^{0}C - 21.45^{0}C = 10.05^{0}C$

q = $151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J$

Moles of NaOH = $101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol$ NaOH

Moles of $H_{2}SO_{4}$ = $50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}$

Enthalpy of the reaction = $\frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol$

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2 years ago

Which event would be impossible to explain by using John Dalton’s model of the atom?

geniusboy [140]

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3 years ago

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The volume of carbon gas at 5.0 ATM was measured to be 363 ml. What will the pressure be if the volume is change to .00020 ml?

aleksley [76]

Answer:

The answer to your question is P2 = 9075000 atm

Explanation:

Data

Pressure 1 = P1 = 5 atm

Volume 1 = V1 = 363 ml

Pressure 2 = P2 = ?

Volume 2 = 0.0002 ml

Process

To solve this problem use Boyle's law

P1V1 = P2V2

-Solve for P2

P2 = P1V1/V2

-Substitution

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P2 = 1815 / 0.0002

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P2 = 9075000 atm

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