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nlexa [21]
3 years ago
7

What substance would you add to nahco3(aq) to form a buffer solution?

Chemistry
2 answers:
Nuetrik [128]3 years ago
8 0

Answer is: H₂CO₃.

This buffer is example of weak acid (H₂CO₃) and its conjugate base (HCO₃⁻).

Dissociation of carbonic acid: H₂CO₃(aq) ⇄ HCO₃⁻(aq) + H⁺(aq).

Adding acid: HCO₃⁻(aq) + H⁺(aq ⇄ H₂CO₃(aq).

A buffer can be defined as a substance that prevents the pH of a solution from changing by either releasing or absorbing H⁺ in a solution.  

Buffer is a solution that can resist pH change upon the addition of an acidic or basic components and it is able to neutralize small amounts of added acid or base, pH of the solution is relatively stable.  

ira [324]3 years ago
7 0
Buffer solution is made by mixing a weak acid with its conjugate base or weak base with its conjugate acid. NaHCO₃(aq) is a conjugate salt and also a conjugate base of H₂CO₃(aq) acid. Hence H₂CO₃(aq) should be added to make the buffer solution.
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What is the mole ratio of N2 to H2 to NH3?
Alexeev081 [22]

Answer:

the answer is 1:3:2

Hope this helps, let me know if you need any other help, Stoichiometry is hard

4 0
3 years ago
If 5.85 grams of cobalt metal react with 15.8 grams of silver nitrate, how many grams of silver metal can be formed and how many
vladimir2022 [97]
Answers:
<span>Answer 1: 10.03 g of siver metal can be formed.</span>
Answer 2: 3.11 g of Co are left over.

Work:

1) Unbalanced chemical equation (given):

<span>Co + AgNO3 → Co(NO3)2 + Ag

2) Balanced chemical equation
</span>
<span>Co + 2AgNO3 → Co(NO3)2 + 2Ag

3) mole ratios

1 mol Co : 2 mole AgNO3 : 1 mol Co(NO3)2 : 2 mol Ag

4) Convert the masses in grams of the reactants into number of moles

4.1) 5.85 grams of Co

# moles = mass in grams / atomic mass

atomic mass of Co = 58.933 g/mol

# moles Co = 5.85 g / 58.933 g/mol = 0.0993 mol

4.2) 15.8 grams of Ag(NO3)

# moles Ag(NO3) = mass in grams / molar mass

molar mass AgNO3 = 169.87 g/mol

# moles Ag(NO3) = 15.8 g / 169.87 g/mol = 0.0930 mol

5) Limiting reactant

Given the mole ratio 1 mol Co : 2 mol Ag(NO3) you can conclude that there is not enough Ag(NO3) to make all the Co react.

That means that Ag(NO3) is the limiting reactant, which means that it will be consumed completely, whilce Co is the excess reactant.

6) Product formed.

Use this proportion:

2 mol Ag(NO3)           0.0930mol Ag(NO3)    
--------------------- =      ---------------------------
    2 mol Ag                              x

=> x = 0.0930 mol

Convert 0.0930 mol Ag to grams:

mass Ag = # moles * atomic mass = 0.0930 mol * 107.868 g/mol = 10.03 g

Answer 1: 10.03 g of siver metal can be formed.

6) Excess reactant left over

    1 mol Co                             x
----------------------- =  ----------------------------
2 mole Ag(NO3)       0.0930 mol Ag(NO3)

=> x = 0.0930 / 2 mol Co = 0.0465 mol Co reacted

Excess = 0.0993 mol - 0.0465 mol = 0.0528 mol

Convert to grams:

0.0528 mol * 58.933 g/mol = 3.11 g

Answer 2: 3.11 g of Co are left over.
</span>


8 0
3 years ago
Which of the following solutions would show the greatest conductivity at 30°C?
Nataly_w [17]
Your answer will be A
7 0
3 years ago
Match each term to its description. (3 points)
madreJ [45]

Answer:   Limiting reactant = 3  

                Theoretical Yield= 1

                  Excess reactant=2

Explanation:  The theoretical yield is the maximum possible mass of a product that can be made in a chemical reaction. It can be calculated from: the balanced chemical equation. the mass and relative formula mass of the limiting reactant , and. the relative formula mass of the product.

An excess reactant is a reactant present in an amount in excess of that required to combine with all of the limiting reactant. It follows that an excess reactant is one remaining in the reaction mixture once all the limiting reactant is consumed.

The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. From the reaction stoichiometry, the exact amount of reactant needed to react with another element can be calculated

3 0
3 years ago
Read 2 more answers
The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k
Delicious77 [7]

Answer:

The vapor pressure at 60.21°C is 327 mmHg.

Explanation:

Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})

We have given in the question

P_1=102\ mmHg

T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol

And R is the Universal Gas Constant.

R=0.008 314 kJ/Kmol

ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165

Taking inverse log both side we get,

\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg

8 0
3 years ago
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