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11111nata11111 [884]
3 years ago
6

Silver nitrate solution reacts with calcium chloride solution according to the equation: 2 AgNO3 + CaCl2 → Ca(NO3)2 + 2 AgCl All

of the substances involved in this reaction are soluble in water except silver chloride, AgCl, which forms a solid (precipitate) at the bottom of the flask. Suppose we mix together a solution containing 6.97 g of AgNO3 and one containing 6.39 g of CaCl2. What mass, in grams, of AgCl is formed?
Chemistry
1 answer:
HACTEHA [7]3 years ago
5 0

Answer:

14.33 g

Explanation:

Solve this problem based on the stoichiometry of the reaction.

To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and  finally calculate the mass of AgCl.

                      2 AgNO₃   + CaCl₂    ⇒   Ca(NO₃)₂   + 2 AgCl

mass, g                  6.97        6.39                                     ?

MW ,g/mol         169.87      110.98                                  143.32

mol =m/MW           0.10         0.06                                    0.10

From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :

0.10 mol x 143.32 g/mol = 14.33 g

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Silver nitrate solution reacts with calcium chloride solution according to the equation: 2 AgNO3 + CaCl2 → Ca(NO3)2 + 2 AgCl All
HACTEHA [7]

Answer:

14.33 g

Explanation:

Solve this problem based on the stoichiometry of the reaction.

To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and  finally calculate the mass of AgCl.

                      2 AgNO₃   + CaCl₂    ⇒   Ca(NO₃)₂   + 2 AgCl

mass, g                  6.97        6.39                                     ?

MW ,g/mol         169.87      110.98                                  143.32

mol =m/MW           0.10         0.06                                    0.10

From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :

0.10 mol x 143.32 g/mol = 14.33 g

5 0
3 years ago
Question 2Classify the following organic structures:CH3OCH3CH2CH2CHCH2CH3CH3NHCH3CH3CH(OH)CH₂CH3
Naily [24]

Within the options, we have four organic compounds. Let's see what the skeletal structure of the compounds is in order to identify them better:

The first compound CH3OCH3 has two methyl groups linked by a carbon atom, this type of compound is called an Ether

The second compound has a double bond, it is badly written but it seems that is an alkene.

The third compound has two methyl groups linked by nitrogen atoms, therefore will be an amine.

The last compound has a hydroxyl group, therefore it is an alcohol

Answer:

CH3OCH3 Ether

CH2CH2CHCH2CH3 Alkene

CH3NHCH3 Amine

CH3CH(OH)CH₂CH3 Alcohol

3 0
1 year ago
Why do adults have stem cells
Eva8 [605]

Adults have stem cells because they are needed to repair certain tissues in the body such as in the muscles, bone marrow, skin, teeth, heart, and liver, to name a few. Thus, they can develop into many different cell types and function to replace older and damaged tissues.

3 0
3 years ago
Read 2 more answers
A geometric isomer with two alkyl groups on the same side of the carbon-carbon double bond is called
grandymaker [24]
Answer:
             <span>A geometric isomer with two alkyl groups on the same side of the carbon-carbon double bond is called <em>cis</em> Isomer.

Explanation:
                   Geometric isomerism takes place about the double bond in alkenes when the alkyl groups are either situated at the same side (<em>cis</em>) or are situated opposite (<em>trans</em>) to each other.

Example:
               <em>cis</em>-2-Butene (highlighted red)

               <em>trans</em>-2-Butene (highlighted blue)</span>

5 0
3 years ago
Given the balanced equation representing a reaction: 2na(s) + cl2(g) → 2nacl(s) + energy if 46 grams of na and 71 grams of cl2 r
igor_vitrenko [27]
Answer is: 2) 117g.
2Na + Cl₂ → 2NaCl
Step 1: calculate amount of substance of sodium and chlorine.
n(Na) = m(Na)÷M(Na) = 46g ÷ 23 g/mol = 2 mol.
n(Cl₂) = m(Cl₂)÷M(Cl₂) = 71g ÷ 71 g/mol = 1 mol.
Step 2: calculate amount of substance and mass of sodium-chloride.
Because both sodium and chlorine react completely, we can use both n to compare with n of NaCl.
n(Na) : n(NaCl) = 2:2, 2 mol : n(NaCl) = 2:2
n(NaCl) = 2mol, m(NaCl) = 2mol ·5805 g/mol = 117 g.

6 0
3 years ago
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