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pav-90 [236]
1 year ago
15

Thsof is which acid H2S04

Chemistry
1 answer:
Crank1 year ago
6 0

Explanation:

H2So4=sulphuric acid , strong acid

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What is the pH of a solution made by mixing 200 mL of 0.025M HCl and 150 mL of 0.050 M HCl?
seraphim [82]

Answer:

The answer to your question is pH = 1.45  

Explanation:

Data

pH = ?

Volume 1 = 200 ml

[HCl] 1 = 0.025 M

Volume 2 = 150 ml

[HCl] 2 = 0.050 M

Process

1.- Calculate the number of moles of each solution

Solution 1

                Molarity = moles / volume

-Solve for moles

                moles = 0.025 x 0.2

result

               moles = 0.005

Solution 2

               moles = 0.050 x 0.15

-result

                moles = 0.0075

2.- Sum up the number of moles

Total moles = 0.005 + 0.0075

                   = 0.0125

3.- Sum up the volume

total volume = 200 + 150

                     350 ml or 0.35 l

4.- Calculate the final concentration

Molarity = 0.0125 / 0.35

              = 0.0357

5.- Calculate the pH

pH = -log [H⁺]

-Substitution

pH = -log[0.0357]

-Result

pH = 1.45    

8 0
3 years ago
Why can characteristics properties be used to identify unknown substances
frozen [14]

Characteristic properties are used because the sample size and the shape of the substance does not matter.

5 0
3 years ago
Elements in group 2 are called alkaline earth metals.what is most similar about alkaline earth metals
Shtirlitz [24]

Answer:

they are divalent elements

3 0
3 years ago
Which one of the following Acids is most likely to hurt you? Acid A: 1.0 M; Acid B: 0.0001M
vaieri [72.5K]
Acid A, assuming the two acids have the same pH. The M stands for molarity which is how concentrated a substance is (basically the higher the molarity the more concentrated the acid is). However, pH refers to how acidic a substance is. If the two acids have different levels of acidity, the answer may be different.
7 0
3 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
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