Write out a balanced, molecular equation, total ionic equation, and net ionic equation for each:

What causes objects sliding on a surface to stop?

MAXImum [283]

Answer:c

Explanation: becasue if u add to it u get 5 or c

8 0

3 years ago

How many moles of electrons are required to reduce one mole of oxygen gas (O2) to two moles of oxygen ions (O2-)?

sergey [27]

O₂ + 4e⁻ ------> 2O²⁻

:)

7 0

3 years ago

Read 2 more answers

NH4NO3 + Na3PO4 → (NH4)3PO4 + NaNO3

Papessa [141]

Based on the equation of the reaction and the data provided,

NH4NO3 is the limiting reactant
mass of Na3PO4 left is 29.375 g
18.75 g of (NH4)3PO4 is produced
31.875 g of NaNO3 is produced

<h3>What are limiting reactants?</h3>

A limiting reactant is a reactant which is used up in a reaction after which the reaction stops.

In the given reaction:

3 NH4NO3 + Na3PO4 → (NH4)3PO4 + 3 NaNO3

The limiting reactant is determined from the stoichiometry of the eqaution.

Moles of reactant = mass/molar mass

Molar mass of NH4NO3 = 80 g/mol

Molar mass of Na3PO4 = 165 g/mol

Molar mass of (NH4)3PO4 = 150 g/mol

Molar mass of NaNO3 = 85 g/mol

From the equation of the reaction, 240 g (3 × 80) of NH4NO3 is required to react with 165 g of Na3PO4

There are only 30.0 g of NH4NO3 reacting with 50.0 g of Na3PO4

30 g of Na3PO4 will react with 30 × 165/240 = 20.625 g of Na3PO4

Therefore, NH4NO3 is the limiting reactant

Na3PO4 is the excess reactant

mass of Na3PO4 left = 50 - 20.625

mass of excess reactant left = 29.375 g

moles of NH4NO3 in 30 g = 30/80 = 0.375 moles

3 moles of NH4NO3 produces 1 mole of (NH4)3PO4

0.375 moles of NH4NO3 will produce 0.375 × 1/3 = 0.125 moles of (NH4)3PO4

mass of 0.125 moles of (NH4)3PO4 = 0.125 × 150

mass of (NH4)3PO4 produced = 18.75 g of (NH4)3PO4

3 moles of NH4NO3 produces 3 moles of NaNO3

0.375 moles of NH4NO3 will produce 0.375 moles of NaNO3

mass of 0.375 moles of NaNO3 = 0.375 × 85

mass of NaNO3 produced = 31.875 g of NaNO3

Learn more about limiting reactant at: brainly.com/question/24945784

4 0

2 years ago

What is the order of the types of nuclear radiation from lowest to highest energy

djverab [1.8K]

Answer:

types include, in order of increasing frequency, radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays and gamma rays.

7 0

4 years ago

Determine whether each of these reactions occur through an S N 1 , S N 2 , E1, or E2 mechanism. A. A tertiary alkyl halide is tr

Elan Coil [88]

Answer:

See explanation below

Explanation:

First to all, we need to remember the difference between Sn and E reactions.

Reactions where a Sn1 or Sn2 occur, it's where you have a reactant with a very good leaving group and a nucleophyle, and the product or products of the reaction always results that the nucleophyle is now taking the place where the leaving group of the innitial reactant was. This can be done in acid medium (SN1 carbocation intermediate 2 steps) or basic medium (SN2 fast reaction 1 step)

Reactions where a E1 or E2 occur, we also have a reactant with a very good leaving group and a nucleophyle, and the product of reaction always result in the fact that not only the leaving group is out of the innitial reactant but also an electrophyle from the reactant is no longer there, and the product form a new pi bond (a double bond or triple bond) and the nucleophile gets the electrophyle that leaves the molecule. Depending of the conditions it can be either E1 or E2 (same conditions as Sn1 and Sn2).

According to this very brief explanation, let's see what we have in the options:

A. Tertiary alkyl halide with ethanol, heating and the product is an alkene.

According to the explanation, we have a tertiary alkyl halide (a carbon with all of it's bond different than hydrogen), reacting with ethanol and heating. The product formed was an alkene. In this case, the halide leaves the molecule forming a carbocation intermediate and the ethanol just react with an electrophyle of the bromide alkyl, in this case, an atom of hydrogen, and then the lone pairs form a double bond giving the alkene. Because of the carbocation intermediate this is a E1 reaction.

B. Tertiary alkyl halide treated with potassium terbutoxide in tert butanol and give an alkene.

This is similar as case A, the difference is that we are doing this in basic medium, therefore, the mechanism involved is either E2 or Sn2. However, the final product is also an alkene, which means that the tert-butoxide substract the electrophyle of the alkyl halide, the lone pairs goes down to the molecule promoving the leaving of the halide to form the alkene. This then, would be an E2 mechanism reaction.

C. is the same as case A so it's a E1 reaction.

3 0

3 years ago