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inysia [295]
3 years ago
7

PLEASE! :( Two aqueous solutions of AgNO3 and NaCl are mixed. Which of the following diagrams best represents the mixture? For s

implicity, water molecules are not shown (Ag + = gray, Cl- = orange, Na + = green, NO ^ - 3 = blue) PLEASE I NEED HELP I ONLY HAVE 15 MINS PLS :'((​

Chemistry
1 answer:
-BARSIC- [3]3 years ago
5 0

Answer:

NaCl + Ag(NO₃) --> Na(NO₃) + AgCl

Explanation:

Chemical equation:

NaCl + Ag(NO₃) --> Na(NO₃) + AgCl

The given reaction represent the double displacement reaction. Anion and cation of both reactant exchange with each others.

The anion of sodium chloride(Cl⁻) combine with cation of silver nitrate Ag⁺ and  NO₃⁻ combine with Na⁺.

The third equation is correct while others are in correct.

Double replacement:

It is the reaction in which two compound exchange their ions and form new compounds.

AB + CD → AC +BD

SORRY CAUSE IDK IF THAT HELPED :(

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How many chloride ions are in 2.6 moles of CaCl2?
Rama09 [41]

Answer:

31.31× 10²³ number of Cl⁻ are present in 2.6 moles of CaCl₂ .

Explanation:

Given data:

Number of moles of CaCl₂ = 2.6 mol

Number of Cl₂ ions = ?

Solution:

CaCl₂  → Ca²⁺ + 2Cl⁻

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

In one mole of CaCl₂ there are two moles of chloride ions present.

In 2.6 mol:

2.6×2 = 5.2 moles

1 mole Cl⁻ =   6.022 × 10²³ number of Cl⁻ ions

5.2 mol ×  6.022 × 10²³ number of Cl⁻  / 1mol

31.31× 10²³ number of Cl⁻

8 0
3 years ago
At sea level, a kilogram weighs approximately:
Marysya12 [62]
The answer would be b, or 2
7 0
3 years ago
Read 2 more answers
For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium car
Alexus [3.1K]

Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • Na₂CO₃: 1 mole
  • Ca(NO₃)₂: 1 mole
  • CaCO₃: 1 mole
  • NaNO₃: 2 mole

Being the molar mass of the compounds:

  • Na₂CO₃: 106 g/mole
  • Ca(NO₃)₂: 164 g/mole  
  • CaCO₃: 100 g/mole
  • NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

  • Na₂CO₃: 1 mole* 106 g/mole= 106 g
  • Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
  • CaCO₃: 1 mole* 100 g/mole= 100 g
  • NaNO₃: 2 mole* 85 g/mole= 170 g

You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of  CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of  CaCO₃?

mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}

mass of CaCO₃= 74.81 grams

<u><em>74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.</em></u>

6 0
3 years ago
How might an increasing population of robins positively and negatively affect a community?
Oliga [24]

All populations of living things are interrelated. When one population of animals, plants, or insects increase or decrease, other populations of living things are also affected. For example, when shrubs and brushy areas are removed from an ecosystem, the rabbit population will likely go down. The reduced rabbit population will lower predator populations that use rabbits as a food source.

In another example, let's assume all the dead hollow trees are removed from a forest ecosystem. Cavity nesting animals such as bluebirds, nuthatch, wrens, screech owls, squirrels and woodpeckers have very little, if any, shelter available. The number of animals of this type would be reduced. Insect populations could increase because of fewer insect eating birds and trees and other plants could be negatively affected. The whole ecosystem is affected.

The amount of suitable habitat for a species of wildlife will determine the number of animals that can survive in the area. Human activity has the greatest impact on the amount and quality of wildlife habitat in Illinois. Wildlife habitat can be destroyed or its quality diminished as a result of urban sprawl, agricultural practices, pollution, sedimentation, or habitat fragmentation.

People can also have a positive impact on wildlife populations through improvement and protection of habitat or ecosystems. The planting of trees and shrubs, as well as wildlife food plots, in the appropriate locations is one way landowners can improve wildlife habitat. People can protect ponds, streams, rivers and wetlands from sedimentation by reducing soil erosion on lands surrounding these aquatic ecosystems. Nesting boxes placed in ecosystems that lack dead, hollow trees will enhance the habitat for cavity nesting animals. There are many things people can do to improve habitat for wildlife.

There are a number of natural resource management agencies in Illinois that can help people enhance and protect wildlife habitat. Listed below are a few of the public agencies which can provide ecosystem management expertise to people in Illinois.

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5 0
3 years ago
Gaseous ICl (0.20 mol) was added to a 2.0 L flask and allowed to decompose at a high temperature:
Ne4ueva [31]

Answer:

The Kc is 1.36 (but this is not an option, may be the options are wrong, or may be I was .. Thanks!)

Explanation:

Let's think all the situation.

               2 ICl(g)   ⇄   I₂(g)    +    Cl₂(g)

Initially      0.20              -               -

Initially I have only 0.20 moles of reactant, and nothing of products. In the reaction, an x amount of compound has reacted.

React          x              x/2               x/2

Because the ratio is 2:1, in the reaction I have the half of moles.

So in equilibrium I will have

           (0.20 - x)          x/2             x/2

Notice that I have the concentration in equilibrium so:

0.20 - x = 0.060

x = 0.14

So in equilibrium I have formed 0.14/2 moles of I₂ and H₂ (0.07 moles)

Finally, we have to make, the expression for Kc and remember that must to be with concentration in M (mol/L).

As we have a volume of 2L, the values must be /2

Kc = ([I₂]/2 . [H₂]/2) / ([ICl]/2)²

Kc = (0.07/2 . 0.07/2) / (0.060/2)²

Kc = 1.225x10⁻³ / 9x10⁻⁴

Kc = 1.36

8 0
3 years ago
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