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JulijaS [17]
3 years ago
11

Do bases contain hydrogen ions (H+)

Chemistry
1 answer:
Dmitrij [34]3 years ago
6 0

One definition of an acid is a molecule that can donate a H+ ion. So for example if you have an acid such as hydrochloric acid (HCl) in water, HCl will donate it's H+ to the H2O molecule, forming Cl- and H3O+.  

In contrast, a base is a molecule that accepts H+ ions. This means a base such as NH3 reacts with H2O to form OH- and NH4.

So when pH is measured, it is the ratio of H3O+ ions to OH- ions. If there are more H3O+ ions, you then know that there are more acidic molecules in the solution, since they must be giving away their H+ ions.  

So a solution with an acid in it will have more H3O+ ions (which can be described as H+ ions) in it. However, an acid by itself will have nothing to donate H+ ions to.

hope this helps also please make me the brainliest

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How much energy (in Joules) is required to convert 129 grams of ice at −23.0 °C to liquid water at 18.0 °C?
Karo-lina-s [1.5K]

Answer:

The energy that is required for the process is:

6230.7 J + 42957 J + 9715.2 J = <u>58902.9 joules</u>

Explanation:

This is a calorimetry problem:

Q = m . C . ΔT

Q = heat; m = mas; C is the specific heat and

ΔT = Final T° - Initial T°

Q = C lat . m

Q = Heat

m = mass

C lar = Latent heat of fusion

First of all we calculate the heat for ice, before it takes the melting point. (from -23°C  to 0°C)

Q = 129 g . 2.10 J/g°C . (0°C - (-23°C)

Q = 129 g . 2.10 J/g°C . 23°C → 6230.7 joules

Then, the ice has melted. To be melted and change the state it required:

Q = C lat . m

Q = 333 J/°C . 129 g → 42957 joules

And in the end, we have water that changed its T° from O°C to 18°C

Q = 129 g . 4.184 J/g °C . (18°C - 0°C)

Q = 9715.2 Joules

The energy that is required for the process is:

6230.7 J + 42957 J + 9715.2 J = 58902.9 joules

5 0
3 years ago
Jf,fyu,eudyj,yudu.kjm,fj
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Answer:

Yes that is correct

Explanation:

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2 years ago
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HELP PLEASE I HAVE A TEST TODAY AND I DON'T UNDERSTAND ANY OF THIS...
myrzilka [38]

Answer:

About 67 grams or 67.39 grams

Explanation:

First you would have to remember a few things:

 enthalpy to melt ice is called enthalpy of fusion.  this value is 6.02kJ/mol

  of ice  

 it takes 4.18 joules to raise 1 gram of liquid water 1 degree C

 water boils at 100 degrees C and water melts above 0 degrees C

 1 kilojoules is 1000 joules

  water's enthalpy of vaporization (steam) is 40.68 kJ/mol

  a mole of water is 18.02 grams

  we also have to assume the ice is at 0 degrees C

Step 1

Now start with your ice.  The enthalpy of fusion for ice is calculated with this formula:

q = n x ΔH    q= energy, n = moles of water, ΔH=enthalpy of fusion

Calculate how many moles of ice you have:

150g x (1 mol / 18.02 g) = 8.32 moles

Put that into the equation:

q = 8.32 mol x 6.02 = 50.09 kJ of energy to melt 150g of ice

Step 2

To raise 1 gram of water to the boiling point, it would take 4.18 joules times 100 (degrees C)  or 418 joules.

So if it takes 418 joules for just 1 gram of water, it would take 150 times that amount to raise 150g to 100 degrees C.  418 x 150 = 62,700 joules or 62.7 kilojoules.

So far you have already used 50.09 kJ to melt the ice and another 62.7 kJ to bring the water to boiling.  That's a total of 112.79 kJ.

Step 3

The final step is to see how much energy is left to vaporize the water.

Subtract the energy you used so far from what you were told you have.

265 kJ - 112.79 kJ = 152.21 kJ

Again q = mol x ΔH (vaporization)

You know you only have 152.21 kJ left so find out how many moles that will vaporize.

152.21 kJ = mol x 40.68  or   mol = 152.21 / 40.68  = 3.74 moles

This tells you that you have vaporized 3.74 moles with the energy you have left.

Convert that back to grams.

3.74 mol   x  ( 18.02 g / 1 mol ) = 67.39 grams

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How to round the decimal 4.15 to the hundredths place?​
gladu [14]

Answer:4.15 is already in the hundreths place so you are good!

if it was 4.155 then you would round it to 4.16 but its NOT that so the answer IS 4.15

Explanation:

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the chemical equation will be XY2

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