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lara31 [8.8K]
3 years ago
13

If the pH of an acid is 4.0, what is the pOH?

Chemistry
2 answers:
LiRa [457]3 years ago
6 0
Well from my research I've calculated that pH + pOH = 14.00. in conclusion
pOH = 14.0 - 4.0 and so forth that equals 10.0 your welcome
My name is Ann [436]3 years ago
4 0
Well from research, I've calculated that pH + pOH = 14.00. In the end, pOH = 14.0 - 4.0 and so forth that equals 10.0.
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Given that NASA needs 151,200 g of Oxygen (O2) per person for a mission, how many grams of water (H2O) does NASA need to ship to
IgorC [24]

this is so ez,

ok so here what u'll do after u balance the equation:

1. convert grams of oxygen to moles of oxygen.

2. convert moles of oxygen to moles of water

3. convert moles of water to grams of water.

4. bOOm... that's your answer

as follows:

2H2O(l)===>2H2(g)+O2(g)

1. 151200(g)/16(g/mol)=9540 moles of O2

2. 9450 moles of O2 × 2 moles of H2O =18900 moles of H2O

3. 18900 moles of H2O × 10g/mol = 189000g of H2O is required.

6 0
4 years ago
When continental plates pull apart at a divergent boundary on land, a(n)____________forms
Alenkinab [10]
Rift/Mid-Ocean Ridge.
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3 0
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Can someone answer all 3 questions plz?
nordsb [41]

Question one should be obvious..

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Question two

No, because even when water is broken down, there should only be one molecule of oxygen. The fact there is two means it's incorrect.

Question three

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3 0
3 years ago
Can you give me an example of a chemical change?
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8 0
3 years ago
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Calculate the amount of time required to raise temperature of 0.50 liter of water from
Burka [1]
<h3>Answer:</h3>

19 seconds

<h3>Explanation:</h3>

Heat energy was supplied and used to heat 0.50 liters of water from 0°C to 10°C.

This means water gained heat energy that was supplied.

Therefore;

Heat supplied = Heat gained by water

We are required to calculate the time taken to raise the temperature of water from 0°C to 10°C.

Step 1: Calculate the heat gained by water

Quantity of heat = Mass × specific heat capacity × change in temperature

Density of water 1 g/ml

Volume of water is 500 mL

Therefore, since Mass = density × volume

The mass of water = 500 g

Change in temperature = 10° C

Specific heat capacity of water = 4.18 J/g/°C

Thus;

Quantity of heat = 500 g × 4.18 J/g/°C × 10

                           = 20,900 Joules

Step 2: Heat supplied by a heater

Heat supplied = Power × time

Power = 1100 J/s

Assuming the time required is x

Heat supplied = 1100x Joules

Step 3: Time required

Remember; heat supplied = heat gained by water

Therefore;

1100x joules = 20,900 Joules

      x = 20,900/1100

        = 19 seconds

Therefore, the time required to raise the temperature of water from 0°C to 10°C is 19 seconds.

6 0
3 years ago
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