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Whitepunk [10]
3 years ago
11

Which of the following statements are TRUE of buffer solutions? 1. A buffer solution can be made by mixing equal concentrations

of ​ acetic acid and sodium acetate. 2. A buffer solution can be made by mixing equal concentrations of ​ hydrochloric acid and sodium chloride. 3. A buffer solution resists changes in pH when small quantities of acid ​ or base are added.
Chemistry
1 answer:
Leokris [45]3 years ago
5 0

Answer:

Option 1 and 3 are correct.

Explanation:

Buffer solution is the solution which resists the change in the magnitude of the pH when small additions of either acid or base is added.  Buffer solutions consist of weak acid and its conjugate base usually which are mixed in relatively equal and large quantities.

Hence,

Option 1. is correct because acetic acid is a weak acid and sodium acetate is the conjugate base of it which are mixed in equal concentrations.

Option 2. is incorrect because HCl and NaCl are both strong acid and salt respectively and will not form a buffer.

Option 3. is correct which is the definition of a buffer solution.

Option 1 and 3 are correct.

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Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti
koban [17]

Answer:

1) 2.0 g

2) 0 g

3) 4.17 g

4) 2.57 g

Explanation:

First of all, we need to know the compounds and the reaction. The ion carbonate is CO3^{-2}, and the ion nitrate is NO3^{-}.

Sodium is in group 1, so it must lose one electron to be stable, and be the cation Na^{+}. Silver has only one electron too, so the cation will be Ag^{+}.

To form the chemical compounds, first we put the cation, then the anion, and change their charges without the signal:

Sodium carbonate: Na2CO3

Silver nitrate: AgNO3

Silver carbonate: Ag2CO3

Sodium nitrate: NaNO3

The balanced reaction will be:

Na2CO3 + 2 AgNO3 --> Ag2CO3 + 2 NaNO3

Now, we must check the stoichiometry, which will be 1:2:1:2 (always in number of moles)

The question wants to know the mass value, so we need to know the molar mass of these compounds. Checking the periodic table will see that:

Na = 23 g/mol, C = 12 g/mol, N = 14 g/mol, O = 16 g/mol, Ag = 108 g/mol

So the molar mass of the compounds must be:

Na2CO3 = 106 g/mol (2x23 + 12 + 3x16)

AgNO3 = 170 g/mol (108 + 14 + 3x16)

Ag2CO3 = 276 g/mol (2x108 + 12 + 3x16)

NaNO3 = 85 g/mol

We have a mixture of the reactants, so one probably would be in excess, so, first will need to test. Let's do the stoichiometry calculus using silver nitrate as the limit, so:

1 mol of Na2CO3 ---------- 2 mol of AgNO3

106 g ------------------------------ 2x170 = 340 g

x ------------------------------------ 5.14 g

By a simple direct three rule:

340x = 544.84

x = 1.6 g of Na2CO3

That means that for this reaction, we only need 1.6 g of Na2CO3 to react with 5.14 of AgNO3. How we have 3.60 g of Na2CO3, it is on excess, and all the AgNO3 will be consumed.

1) The mass of Na2CO3 that remains after the reaction will be the initial less the mass that reacted:

m = 3.6 - 1. 6 = 2.0 g

2) All the AgNO3 reacted, so there isn't a mass present after the reaction.

m = 0 g

3) Now, doing the stoichiometry calculus between AgNO3 and Ag2CO3

2 moles of AgNO3 ------------- 1 mol of Ag2CO3

2x170 g ------------------------------- 276 g

5.14 g --------------------------------- x

By a simple direct three rule:

340x = 1418.64

x = 4.17 g of Ag2CO3

4) Now, doing the stoichiometry calculus between AgNO3 and NaNO3

2 moles of AgNO3 ----------------------- 2 moles of NaNO3

2x170 g ---------------------------------------- 2x85 g

5.14 g ------------------------------------------- x

By a simple direct three rule:

340x = 873.8

x = 2.57 g

8 0
3 years ago
A hot lump of 39.9 g of iron at an initial temperature of 78.1 °C is placed in 50.0 mL H 2 O initially at 25.0 °C and allowed to
Drupady [299]

Answer : The final temperature of the mixture is 29.6^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron = 0.499J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of iron = 39.9 g

m_2 = mass of water  = Density\times Volume=1g/mL\times 50.0mL=50.0g

T_f = final temperature of mixture = ?

T_1 = initial temperature of iron = 78.1^oC

T_2 = initial temperature of water = 25.0^oC

Now put all the given values in the above formula, we get

(39.9g)\times (0.499J/g^oC)\times (T_f-78.1)^oC=-(50.0g)\times 4.18J/g^oC\times (T_f-25.0)^oC

T_f=29.6^oC

Therefore, the final temperature of the mixture is 29.6^oC

8 0
3 years ago
The rate of effusion of oxygen to an unknown gas is 0.935. what is the other gas?
Kamila [148]

Answer:

N2

Explanation:

Rate of effusion is defined by Graham's Law:  

(Rate 1/Rate 2) = (sqrt (M2)/ sqrt (M1))

(Where M is the molar mass of each substance. )

Molar Mass of oxygen, O2, is 32 (M1).

Rate of effusion of O2 to an unknown gas is .935(Rate 1).  

Rate 2 is unknown so put 1.

Solve for x (M2).

.935/1 = sqrt x/ sqrt32

.935 x sqrt 32 = sqrt x

5.29 = sq rt x

5.29^2 = 27.975 = 28  

N2 has a molar mass of 28 so it is the correct gas.

4 0
3 years ago
Consider the chemical equation below.
Aleks [24]
I would say A but then again im not too sure so hope that makes it easier to somehow
6 0
3 years ago
Read 2 more answers
What would be the formula of the precipitate that forms when pb(no3)2 (aq) and k2so4 (aq) are mixed?
maria [59]
The formula of the ppt. formed is PbSo4 , which is inslouble.
6 0
3 years ago
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