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Ierofanga [76]
2 years ago
8

Write balanced, net ionic equations for the following precipitation, acidn base, or gasn forming reactions. Include states of ma

tter. Hint: You should write the molecular equation first to predict the products. a) mixing aqueous solutions of iron (III) chloride and lithium sulfide b) mixing aqueous solutions of sodium acetate and ammonium phosphate c) mixing aqueous solutions of perchloric acid and potassium hydroxide d) mixing aqueous solutions of ammonia and nitric acid e) mixing aqueous solutions of nitrous acid and sodium hydroxide f) adding aqueous hydroiodic acid to solid calcium carbonate g) Problems c) through f) are acid- base reactions. Comment on the differences in the net ionic equations for these reactions.
Chemistry
1 answer:
dybincka [34]2 years ago
5 0

Answer:

Explanation:

a ) 2FeCl₃ + 3Li₂S  = Fe₂S₃ ( s )  + 6 LiCl

2Fe⁺³ + 6Li ⁻ + 6Cl⁻ + 3S⁻² = 6Li + 6Cl⁻ + Fe₂S₃ ( s )

b  )

3CH₃COONa +( NH₄)₃PO₄ = 3CH₃COONH₄ + Na₃PO₄

3CH₃COO + 3Na⁺ +  3NH₄⁻  +  PO₄⁺³  = 3CH₃COO⁻ +3NH₄⁺ + Na₃PO₄

c )

HClO₄  + KOH = kClO₄ + H₂O

H ⁺ + ClO₄⁻ +  K⁺ + OH⁻  =  k⁺  ClO₄⁻  + H₂O

d )

NH₄OH + HNO₃ = NH₄NO₃  + H₂O

NH₄⁺ + OH⁻ + H⁺ + NO₃⁻ = NH₄⁺ + NO₃⁻ +  H₂O

e )

HNO₂ + KOH = KNO₂ + H₂O

H⁺ + NO₂⁻ + K⁺ + OH⁻ = K⁺ + NO₂⁻ + H₂O

f ) HIO₃ + CaCO₃ ( s ) = Ca( IO₃ )₂ + H₂CO₃

H⁺ + IO₃⁻ + CaCO₃ ( s ) = Ca( IO₃ )₂ + H₂CO₃

g )

c ) is strong acid and strong base

d ) is weak base and strong acid

e ) weak acid and strong base

f ) Strong acid and basic salt

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if a scientist creates a new rock like substance in a laboratory, why wouldnt this type of material be classified as a true mine
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The reason why it is not considered this is because the material was made in a lab, not through nature, which is what is required to be considered as a true mineral.
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3 years ago
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You drop a rock weighing 23.2 g into a graduated cylinder that contains 55 mL. The level of the water rises to 62 mL. What is th
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The density of the rock is 3.314g/mL

CALCULATE DENSITY:

  • According to this question, a rock weighs 23.2g. After dropping the rock into a graduated cylinder containing 55mL of water, the level changes to 62mL.

  • This means that the volume of the rock can be calculated as follows:

Volume of rock = 62mL - 55mL

Volume of rock = 7mL

Density can be calculated using the formula as follows:

Density = mass ÷ volume

Density = 23.2 ÷ 7

Density = 3.314g/mL

Therefore, the density of the rock is 3.314g/mL

Learn more: brainly.com/question/6034174?referrer=searchResults

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2 years ago
What do you think would happen if you had a very small amount of water in a glass and you
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3 years ago
A saturated solution of baso4 has a concentration of 0.5mol/l. a 55ml sample is taken by you. what is the mass of baso4 in the s
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Answer:

6.4 g BaSO₄

Explanation:

You have been given the molarity and the volume of the solution. To find the mass of the solution, you need to (1) find the moles BaSO₄ (via the molarity ratio) and then (2) convert moles BaSO₄ to grams BaSO₄ (via the molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs to reflect the sig figs of the given values.

Molarity (mol/L) = moles / volume (L)

(Step 1)

55 mL / 1,000 = 0.055 L

Molarity = moles / volume                             <----- Molarity ratio

0.5 (mol/L) = moles / 0.055 L                        <----- Insert values

0.0275 = moles                                             <----- Multiply both sides by 0.055

(Step 2)

Molar Mass (BaSO₄): 137.33 g/mol + 32.065 g/mol + 4(15.998 g/mol)

Molar Mass (BaSO₄): 233.387 g/mol

0.0275 moles BaSO₄          233.387 g
---------------------------------  x  -------------------  =  6.4 g BaSO₄
                                                1 mole

6 0
2 years ago
g Phosphorus -32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The h
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Answer:

6.88 mg

Explanation:

Step 1: Calculate the mass of ³²P in 175 mg of Na₃³²PO₄

The mass ratio of Na₃³²PO₄ to ³²P is 148.91:31.97.

175 mg g Na₃³²PO₄ × 31.97 g ³²P/148.91 g Na₃³²PO₄ = 37.6 mg ³²P

Step 2: Calculate the rate constant for the decay of ³²P

The half-life (t1/2) is 14.3 days. We can calculate k using the following expression.

k = ln2/ t1/2 = ln2 / 14.3 d = 0.0485 d⁻¹

Step 3: Calculate the amount of P, given the initial amount (P₀) is 37.6 mg and the time elapsed (t) is 35.0 days

For first-order kinetics, we will use the following expression.

ln P = ln P₀ - k × t

ln P = ln 37.6 mg - 0.0485 d⁻¹ × 35.0 d

P = 6.88 mg

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2 years ago
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