All categories

3 years ago

Which is most likely to happen during a precipitation reaction?

Chemistry

1 answer:

Salsk061 [2.6K]3 years ago

3 0

Answer:

I think its B

Explanation:

Precipitation reactions leave a solid behind. The solid is called a precipitate.

You might be interested in

When 0.875 g. of magnesium was heated in air, 1.25 g. of magnesium oxide

Ilia_Sergeevich [38]

Answer:

1. What is the coefficient for calcium in the balanced equation?

2 Ca + 5 NaOH → 4 Na + 2 Ca(OH)2

2. The burning of 20.0 g of acetylene (C2H2) produces what mass of carbon dioxide? You must

write the equation, balance it and solve the mass-mass problem.

a. Balanced equation:

2 C2H2 + 5 O2  4 CO2 + 2 H2O

Explanation:

3 0

3 years ago

How many molecules are in 2.00 moles of H20????????

Elis [28]

2x 6.022x10^23= 1.204x10^24

7 0

3 years ago

How do I round 10.25 to only 3 significant digits

Lady_Fox [76]

10.3 is good the correct rounding of three sig figures

3 0

3 years ago

The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In t

Sloan [31]

Answer:

Mass = 42.8g

Explanation:

4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )

Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.

Step 1: Determine the balanced chemical equation for the chemical reaction.

The balanced chemical equation is already given.

Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).

Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol

Oxygen = 63.4g × 1mol / 32g = 1.9813mol

Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.

If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.

Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.

5 moles of O2 = 6 moles of H2O

1.9831 moles = x

x = (1.9831 * 6 ) / 5

x = 2.37972 moles

Mass of H2O = Molar mass * Molar mass

Mass = 2.7972 * 18

Mass = 42.8g

6 0

3 years ago

Brainliest for correct answer please show all work

Korvikt [17]

Answer:

1) Na₃PO₄ + 3 KOH ➙ 3 NaOH + K₃PO₄

2) MgF₂ + Li₂CO₃➙ MgCO₃ + 2 LiF

3) P₄ + 3 O₂➙ 2 P₂O₃

Explanation:

To balance an equation, ensure that the number of atoms of each element is equal on both sides.

Reactants would be those on the left of the arrow while products are on the right of the arrow.

Balance O and H atoms last.

Question 1:

__Na₃PO₄ + __KOH ➙ __NaOH + __K₃PO₄

Reactants: 3Na, 1P, 1K, 5O, 1H

Products: 1Na, 1P, 3K, 5O, 1H

Balance the number of Na:

__Na₃PO₄ + __KOH ➙ 3 NaOH + __K₃PO₄

Reactants: 3Na, 1P, 1K, 5O, 1H

Products: 3Na, 1P, 3K, 7O, 3H

Balance the number of K:

__Na₃PO₄ + 3 KOH ➙ 3 NaOH + __K₃PO₄

Reactants: 3Na, 1P, 3K, 7O, 3H

Products: 3Na, 1P, 3K, 7O, 3H

The equation is now balanced.

Question 2:

__MgF₂ + __Li₂CO₃➙ __MgCO₃ + __LiF

Reactants: 1Mg, 2F, 2Li, 1C, 3O

Products: 1Mg, 1F, 1Li, 1C, 3O

Balance number of Li and F atoms:

__MgF₂ + __Li₂CO₃➙ __MgCO₃ + 2 LiF

Reactants: 1Mg, 2F, 2Li, 1C, 3O

Products: 1Mg, 2F, 2Li, 1C, 3O

The equation is now balanced.

Question 3:

__P₄ + __O₂➙ __P₂O₃

Reactants: 4P, 2O

Products: 2P, 3O

Balance the number of P atoms:

__P₄ + __O₂➙ 2 P₂O₃

Reactants: 4P, 2O

Products: 4P, 6O

Balance the number of O atoms:

__P₄ + 3 O₂➙ 2 P₂O₃

Reactants: 4P, 6O

Products: 4P, 6O

The equation is now balanced.

4 0

3 years ago