List three possible results of a double replacement reaction

Answers

Answer 1
Answer: 1. two compounds form
2. precipitate forms
3. ions exchanged between the two reactants

Related Questions

Enter the molecular formula for butane, C4H10express your answer as a chemical formula
What is the purpose of interphase in the eukaryotic cell cycle?
In lower concentrations, salts generally _____. dissolve precipitate liquefy solidify
Which biogeochemical cycle is the only one that does not include the atmosphere?
If it takes 65.3 min for the concentration of a reactant to drop to 20.0% of its initial value in a first-order reaction, what is the rate constant for the reaction in the units min-1?

Martin uses 5/8 of a gallon of paint to cover4/5 of a wall. What is the unit rate at which Martin paints in walls per gallon

Answers

Answer:

Unit rate at which Martin paints in walls per gallon is (32)/(25) wall/gallon.

Explanation:

Rate is defined as ratio between two quantities with different units.

Gallons of paint used by Martin = (5)/(8) gallons

Portion of wall Martin wishes to pint = (4)/(5) wall

Rate at which Martin paints the wall:

Rate=\frac{\text{Portion of wall covered}}{\text{Paint used}}

Rate=((4)/(5) wall)/((5)/(8) gallon)=(32)/(25) wall/gallon

Unit rate at which Martin paints in walls per gallon is (32)/(25) wall/gallon.

What is the ph at the equivalence point when 0.10 m hno3 is used to titrate a volume of solution containing 0.30 g of koh?

Answers

Answer;

pH = 7  

Explanation;

  • HNO3 is a strong acid. KOH is a strong base. At the equivalence point, the solution will be neutral.
  • Therefore; the pH at the equivalence point when 0.10 m HNO3 is used to titrate a volume of solution containing 0.30 g of KOH will be 7.
  • The neutralization of a strong base with a strong acid gives a salt solution at the equivalence point. The salt contains ions that do not change the pH of water.

Answer:

pH at equivalence point will be 7

Explanation:

  • HNO_(3) is a strong acid and KOH is a strong base. Hence a neutral salt would appear from acid-base reaction between them.
  • Acid-base reaction: HNO_(3)+KOH\rightarrow KNO_(3)+H_(2)O
  • At equivalence point, KNO_(3) and water exists in medium.
  • As KNO_(3) is a neutral salt therefore pH of medium will be equal to 7 (neutral).
  • So, pH at equivalence point will be 7.

the formula C=5/9 (F - 32) can be used to convert a temperature from degrees Fahrenheit to degrees Celsius. which of the following is closest In value to 5/9?

Answers

F = (9)/(5) c +32

Enter the temperature into the degrees Celsius (C) equal to 113∘ Fahrenheit (F). The formula F = 1.8°C + 32 allows you to find the Fahrenheit (F) temperature for a given Celsius (C) temperature.

Given: C = \frac{5}9} (F - 32)

First, multiply both sides by (9)/(5) to get:

(9)/(5) c = F - 32

Add  32 to both sides and transpose to get:

F = (9)/(5)c + 32

These are the formulas to translate between Fahrenheit and Celsius temperatures.

Fahrenheit and Celsius meet at −40 degrees.

That is −40∘ F = −40∘C

So alternative formulas for the conversions can be written:

F = (9)/(5) (C + 40) − 40

C = (5)/(9) (F +40) − 40

Learn more about Fahrenheit and Celsius here brainly.com/question/12897775

#SPJ2

F= 1.8C+32 or C= F-32/1.8

To what volume in millimeters must 50.0 mL of 18.0 M H2SO4 be diluted to obtain 4.35 M H2SO4?

Answers

We know that to relate solutions of with the factors of molarity and volume, we can use the equation: M_(1)  V_(1) = M_(2)  V_(2)

**NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.

So now we can assign values to these variables. Let us say that the 18 M H_(2)  SO_(4) is the left side of the equation. Then we have:

(18 M)(0.050 L)=(4.35M) V_(2)

We can then solve for V_(2):

V_(2)= ((18M)(0.05L))/(4.35M) and V_(2) =0.21 L or 210 mL

We now know that the total amount of volume of the 4.35 M solution will be 210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.


22 Which class of organic compounds contains nitrogen?(I) aldehyde(2) alcohol(3) amine(4) ether

Answers

The class of organic compounds that contains nitrogen is (3) amine.

Amines are organic compounds that contain a nitrogen atom bonded to one or more carbon atoms. They can be considered derivatives of ammonia (NH₃), where one or more hydrogen atoms are replaced by organic groups. Amines play essential roles in biological systems and are involved in various chemical reactions and processes. They can be classified as primary, secondary, or tertiary amines based on the number of carbon groups attached to the nitrogen atom.

Hence, the correct option is an amine option 3.

Learn more about amine here:

brainly.com/question/31391643

#SPJ 6

The answer is (3)amine. The aldehyde structure is -CHO. The alcohol structure is  -OH. The amine structure is -NH2. And the ether structure is R-O-R'. The "R" means other function groups.

A sample of salt (sodium chloride) is placed on the tip of a platinum wire. When the sample is heated on blue flame the flame turned yellow. Explain how this can be done to electrons in sodium atoms

Answers

So platinum is a transition metal. In general transition metals are reducers, which means they can give the electrons they have, to the sodium atoms. Also in chemistry we look at sub orbitals rather that shells(2,8,8). So due to the energy from heat, the d orbital split as electrons move to a higher energy level. Some of the electrons are given to the sodium ions and therefore the flame changes colour to yellow. 
The excitation of the electrons is caused by them getting energy and so moving up an energy level. This energy is released and the electron returns to it's original state. The energy released, however, does not release in the same direction, but in different/various directions. Therefore the colour of the light changes as some energy is released in the surrounding.