I have a few questions from my homework and I was wondering if I can get some help

how the various concentrations of acid will affect the amount of limestone that is dissolved from each jar.

schepotkina [342]

The explanation of the how the various concentrations of acid will affect the amount of limestone has been given below.

Effects of acid rain on limestone:-

When an acid combines with a carbonate, it produces carbon dioxide as a gas and forms a salt that is soluble in the carbonate and acid's water.

There are several gases in the atmosphere that can dissolve in precipitation such as rain and snow.

Some may produce acids in rain water, such as carbonic acid, sulfuric acid, and nitric acid.

Because the concentration is modest, the rain is not highly acidic, but it is acidic enough to react with the carbonates that make up limestone.

Thus we discussed the affects of acid rain on limestones above.

Learn more about Acid Rain here:

brainly.com/question/718250

#SPJ10

8 0

1 year ago

Suppose 550.mmol of electrons must be transported from one side of an electrochemical cell to another in 49.0 minutes. Calculate

nekit [7.7K]

Answer:

18.0 Ampere is the size of electric current that must flow.

Explanation:

Moles of electron , n = 550 mmol = 0.550 mol

1 mmol = 0.001 mol

Number of electrons = N

$N=N_A\times n$

Charge on N electrons : Q

$Q = N\times 1.602\times 10^{-19} C$

Duration of time charge allowed to pass = T = 49.0 min = 49.0 × 60 seconds

1 min = 60 seconds

Size of current : I

$I=\frac{Q}{T}=\frac{N\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}$

$=\frac{n\times N_A\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}$

$I=\frac{0.550 mol\times 6.022\times 10^{23} mol^{-1}\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}=18.047 A\approx 18.0 A$

18.0 Ampere is the size of electric current that must flow.

3 0

3 years ago

Write a balanced overall reaction from these unbalanced half-reactions sn and ag

loris [4]

Balance each one by adding electrons to make the charges on both sides the same:

Sn--> Sn2+ + 2 e-
Ag+ + 1 e- --> Ag

Now, you have to have the same number of electrons in the two half-reactions, so multiply the second one by 2 to get:

2 Ag+ + 2 e- --> 2 Ag

Now, just add the two half reactions together, cancelling anything that's the same on both sides:

2 Ag+ + Sn --> Sn2+ + 2 Ag

And you're done.

8 0

3 years ago

Which one of the atoms shown would be most likely to form a cation with a charge of 1?

olga55 [171]

Ions are stable atoms of elements formed when original atoms loose or gain electrons to form stable configuration. Anions are negatively charged ions formed when atoms gain electrons while cations are the positively charged ions formed when atoms loose electrons. In this case a charge of +1 means an atom lost one electron to form a stable configuration. Therefore, the correct answer is A

6 0

3 years ago

Read 2 more answers

Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of

bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

Step 1: Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

Step 2: The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

Step 3: Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

Step 4: Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

Step 5: Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

Step 6: Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

Step 7: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

Step 8: Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

4 0

3 years ago

​I have a few questions from my homework and I was wondering if I can get some help ​

I have a few questions from my homework and I was wondering if I can get some help