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iris [78.8K]
3 years ago
14

How would the addition of protons affect the concentration of CH3COOH? How would the addition of OH– affect the amount of CH3COO

H present? How would the addition of CH3COO– affect the concentration of protons? What would happen to [H+] if [CH3COOH] were increased?
Chemistry
2 answers:
Deffense [45]3 years ago
8 0

Answer:

1) increase concentration

2) decrease the amount

3) decrease the concentration

4) it would increase

Explanation: edge 2021

fiasKO [112]3 years ago
7 0

Answer:

1) increase concentration

2) decrease the amount

3) decrease the concentration

4) it would increase

Explanation:

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If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi
zlopas [31]

Answer:

T_{eq}=28.9\°C

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

Q_{Cd}+Q_{W}=0

Thus, we insert mass, specific heat and temperatures to obtain:

m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}

Now, we plug in to obtain:

T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0
2 years ago
Which is the correct set of products for the acid-base neutralization reaction between hydrochloric acid and sodium hydroxide?
nydimaria [60]

Answer: I think the answer is C. NaCl and H2O

Explanation: I’m not sure tho

3 0
2 years ago
The Goodyear blimp contains 5.7 x 10^6 L of helium at 25 degrees Celsius and 1 atm. What is the mass in grams of the helium insi
anygoal [31]

Answer:

1.72x10⁻⁵ g

Explanation:

To solve this problem we use the PV=nRT equation, where:

  • P = 1 atm
  • V = 5.7x10⁶ L
  • n = ?
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = 25 °C ⇒ (25+273.16) = 298.16 K

And we <u>solve for n</u>:

  • 1 atm * 5.7x10⁶ L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
  • n = 4.29x10⁻⁶ mol

Finally we <u>convert moles of helium to grams</u>, using its <em>molar mass</em>:

  • 4.29x10⁻⁶ mol * 4 g/mol = 1.72x10⁻⁵ g

8 0
3 years ago
In an chemicalbreaction involving Fe and S, it was found that 45.2 g of Fes was produced. If the percent yield of the reaction i
saw5 [17]

Answer:

47.8 g

Explanation:

Remember the equation for percent yield:

% yield = actual / theoretical

We're given two of the values in the question, so plug n' play:

0.945 = 45.2 / theoretical

theoretical = 47.8 g

Keep in mind you can use mass here without converting to moles because we're working with products only. If you were given a mass of reactants, you would need to convert to moles and using a balanced chemical equation find the corresponding moles of product produced.

7 0
3 years ago
I need help with this!!!
scZoUnD [109]
11- Form of energy that can be reflected or emitted from objects through electrical or magnetic waves.

12-Energy that is caused by moving electric charges.

13-Energy stored in the bonds of chemical compounds.

I only know those. Sorry hoped I helped a little! :)
5 0
3 years ago
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