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Brums [2.3K]
3 years ago
12

1.0 mole of KCl(aq) is added to which solution to produce a precipitate?

Chemistry
1 answer:
taurus [48]3 years ago
7 0

Answer:

Ag+

Explanation:

If you imagine as if the problem were double replacement, you would pair the Cl with one of the following ions provided in the choices. As seen on Table F, Ag+ paired with Cl- produces an insoluble compound, hence the precipitate. All the other ions shown in the multiple choice section, when paired with Cl- will produce a soluble compound, as a result NOT a precipitate.

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Consider the following system at equilibrium where H° = 111 kJ/mol, and Kc = 6.30, at 723 K.
Rashid [163]

Answer:

1) The value of Kc:

C. remains the same.

2) The value of Qc:

A. is greater than Kc.

3) The reaction must:

B. run in the reverse direction to restablish equilibrium.

4) The concentration of N2 will:

B. decrease.

Explanation:

Hello,

In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:

1) The value of Kc:

C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:

ln(K)=\frac{\Delta _RG}{RT}

2) The value of Qc:

A. is greater than Kc, since the reaction quotient is:

Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}

Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.

3) The reaction must:

B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.

4) The concentration of N2 will:

B. decrease, since less reactant is forming the products.

Best regards.

8 0
4 years ago
Read 2 more answers
Chemical reactions that ____ energy will not occur without a source of energy
andrezito [222]
Chemical reactions that release energy will not occur without a source of energy.  So the answer is release.
4 0
3 years ago
The city council of a small town wanted to add fluoride to their water so that all the residents would have healthier teeth. The
azamat

Answer:NaF is ionic. NF3 is covalent. SiF4 is ionic. CaF2 is Ionic and  NH4F is also ionic. Ionic compounds transfer electrons whereas covalent compounds share electrons hence the word "co". Also, ionic compounds are formed with metal and nonmetal. Where a covalent is with 2 nonmetals. Only ionic compounds would produce fluoride in water because ionic compounds can dissolve in water and covalent compounds cant.

3 0
3 years ago
A compound with molecular formula C5H10O2 has the following 1H NMR spectrum: 1.15 (t, 3H), 1.25 (t, 3H), 2.33 (q, 2H), 4.13 (q,
Vilka [71]

Answer:

Ethyl propionate

Explanation:

Calculating the double bond equivalence as:

DBE =  C - (H/2) - (X/2) + (N/2) +1

Where  

C is the number of carbon atoms

N is the number of nitrogen atoms

X is the number of halogens

H is the number of hydrogen atoms

So, according to the formula, C_5H_{10}O_2

DBE =  5 - (10/2) - (0/2) + (0/2) +1 = 1

It means there is one double bond or one ring.

from the NMR signal, it is clear that compound has two triplet and two quardrate group that means two -CH3 and two -CH2 groups are present.

From the splitting information, it is clear that each the -CH2- group is next to a -CH3, and vice-versa. In this case ring structure is not possible.

The chemical shift of two protons 4.13 ppm(q, 2H) is for protons next to an O atom, therefore, compound must have

-OCH_2CH_3.

Then the signal at 2.33 is for a -CH2-group next to a C=O. So compound must have a structure CH3-CH2-C=O. Now, there is only one way to complete the structure - by bonding the O to the C=O carbon.

So the compound is ethyl propionate

3 0
3 years ago
PLEASE HELP!! I REALLY NEED HELP
Allushta [10]

Answer:

Explanation:

1. find the molar mass (amu) of each element and add them to get the whole molar mass.

2. divide the 1 element molar mass with the whole molar mass

3. multiple by 100 and that gives you the % composition.

<h2><u><em>56-57: NaCl</em></u></h2>

1. Na(22.99amu) + Cl (35.453amu)=58.443

2(Na):   \frac{22.99}{58.443} = .393

2(Cl): \frac{35.453}{58.443}= .607

3(Na): .393 * 100=39.3%

3(Cl): .607 * 100= 60.7%

<h2><u>58-60 </u>K_{2} CO_{3}<u /></h2>

1. K: (39.098)(2)=78.196

_ C: (12.011)(1)= 12.011

_O: (15.99)(3) = 47.997

78.196+12.011+47.997= 138.204

2:K: \frac{78.196}{138.204}= .566 <u>Step </u>3: (.566)(100)= 56.6%

2: C: \frac{12.011}{138.204}= .087 <u>Step 3</u>: (.087)(100)= 8.7%

2: O: \frac{47.997}{138.204}= .347 <u>Step 3</u>: (.347)(100) = 34.7%

<h2>61-62 Fe_{3} O_{4}</h2>

1. Fe (55.845)(3)= 167.535

_ O (15.999)(4) = 63.996

167.535+63.996=231.531

2: Fe: \frac{167.535}{231.531}= .724 Step 3: (.724)(100)= 72.4%

2: O : \frac{63.996}{231.531}= .276 Step 3: (.276)(100) = 27.6%

<h2>63-65 C_{3}H_{5}(OH)_{3}</h2>

1.

C(12.011*3)=36.033

H(1.008*5)=5.04 + (1.008*3)=3.024 so its 8.064

O(15.999*3)=47.997

add them: 92.094

2: C: \frac{36.033}{92.094}= .391 Step 3: (.391)(100) = 39.1%

2: H: \frac{8.064}{92.094}= .088 step 3: (.088)(100) = 8.8%

2: O: \frac{47.997}{92.094} = .521 step 3: (.521)(100) = 52.1%

3 0
3 years ago
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