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3 years ago

Three Balanced equations for Ammonium salt+Alkali=Salt+water+carbon dioxide

Chemistry

1 answer:

arlik [135]3 years ago

4 0

Answer:

NH₄Cl (aq) + NaOH (aq) -> NaCl (aq) + H₂O (l) + NH₃ (g)

(NH₄)₂CO₃ (aq) + 2KOH (aq) -> K₂CO₃ (aq) + 2H₂O (l) + 2NH₃(g)

NH₄NO₃ (aq) + LiOH (aq) -> LiNO₃ (aq) + H₂O (l) + NH₃ (g)

Explanation:

The gas liberated when an alkali reacts with an ammonium salt is NH₃ (ammonia), not CO₂.

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When the same or different atoms combine together, you have a:

Alex73 [517]

Answer:

D.) molecule

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3 years ago

Finding the density of solids requires a method of measuring the volume of the solid. If the solid has a regular geometric shape

Eduardwww [97]

Answer:

11 g/cm³

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3 years ago

Octane has a density of 0.692 g/ml at 20oc. how many grams of o2 are required to burn 15.0 gal of c8h18 (the average capacity of

Hatshy [7]

Octane has a density of 0.692 g/ml at 20°C. Grams of O₂ required to burn 15.0 gal of C₈H₁₈ is 1.37868×10⁵g.

Octane is a hydrocarbon having eight carbon atoms and have single bonds only.

Given,

Density of octane = 0.692g/ml

Temperature = 20°C = 293K

Amount of Octane present = 15gal

Molar mass of octane = 114g

We know that,

1 gal = 3.78541L = 3785.4ml

Hence, 15 gal = 56781 ml

Now let's calculate the Mass of octane required:

Mass of octane = 0.692 x 56781

Mass of octane = 39292.45 g

According to the given equation,

C₈H₁₈ + 12.5O₂ ---------> 8CO₂ + 9H₂O

Also, 114g of octane needs-------> 400g of oxygen

39292.45g of octane needs ---------> 137868.24g of oxygen

Hence, oxygen needed to burn octane is 1.37868×10⁵g

Learn more about Octane here, brainly.com/question/21268869

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2 years ago

Which two processes are responsible for water's transfer of transfer of energy in the atmosphere and hydrosphere?

Liula [17]

Oooo! I love this :P
D. Precipitation and evaporation :P
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3 years ago

What is the minimum mass of hubnerite (MnWO4), a tungsten ore, from which 95.0 g of pure tungsten could be extracted?

Gnoma [55]

From one mole of MnWO4, we get one mole of tungsten. Therefore 10.88 mol of tungsten will be obtained from 10.88 mol of MnWO4.

Molar mass MnWO4 = 54.94 + 183.84 + 4(16.0) = 302.78 g.mol-1

Therefore mass of MnWO4 required = mole x molar mass = 10.88 mol x 302.78 g.mol-1 = 3294.24 grams

And, 3294.24 grams = > 3.29424 kg

On rounding off to two significant digits, we get 3.3 kg.

Rounds a number to 3 significant digits the same way it rounds to 3 decimal places. Count up to 3 digits from the first non-zero digit. Then round off the last digit. Pad the remaining digits to the right of the decimal point with zeros. 0.0560 has 3 significant digits. In accordance with scientific conventions, leading zeros in decimal numbers are not treated as significant. So exclude 0.0 from 0.0560.

Learn more about Significant digits here:-brainly.com/question/28052368

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8 0

1 year ago