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PolarNik [594]
3 years ago
6

Calculate the ph of sweater which has a hydrogen ion concentration of 1 times 10 to the power of 8

Chemistry
1 answer:
telo118 [61]3 years ago
5 0

Answer:

The pH of the sweater containing Hydrogen ion concentration

[H^{+}]=1\times 10^{-8} is

<u>8</u>

<u></u>

Explanation:

pH = It is the negative logarithm of activity (concentration) of hydrogen ions.

pH = -log([H+])

Now, In the question the concentration of [H+] ions is :

[H^{+}]=1\times 10^{-8}

pH = -log(1\times 10^{-8})

use the relation:

log(10^{-a})=a

pH= -(-8)

pH = 8

Note : <em><u> 1 times 10 to the power of 8 must be" 1 times 10 to the power of -8"</u></em>

If the concentration is

[H^{+}]=1\times 10^{8}

Then pH = -8 , which is not possible . So in that  case the pH calculation is by other method

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It is 762 miles from here to Chicago. An obese physics teacher jogs at a rate of 5.0 miles every 20.0 minutes. How long would it
Lina20 [59]

Answer:

3,048 minutes

Explanation:

762 divided by 5

that number times 20

5 0
3 years ago
An electromagnetic wave that has a higher frequency than ultraviolet light will have ____ than the ultraviolet light.
8_murik_8 [283]
C)shorter wavelength and higher energy
frequency is inversely proportional to wavelength
frequency is directly proportional to Energy
5 0
3 years ago
What would be the mass in grams of 1.204 x 1024 molecules of sulfur dioxide
melomori [17]

Answer:

mass of sulfur = 96 g

Explanation:

no of moles of sulfur dioxide in 1.204\times 10^{24} molecules = \frac{1.204\times 10^{24}}{avagadro number }= \frac{1.204\times 10^{24}}{6.023\times 10^{23}}

                                       = 2 moles

therefore mass of sulfur dioxide = moles×atomic number

                                                      =2×(16+32)

                                                      =96

6 0
3 years ago
slader An experiment is carried out where 13.9 g of solid NaOH is dissolved in 250.0 g of water in a coffee-cup calorimeter. Dis
kvv77 [185]

Answer:

The mass of the surrounding is M_t = 263.9 \ g

Explanation:

From the question we are told that

      The mass of  NaOH is  m = 13.9 \ g

      The mass of water is m_w = 250.0g

      The chemical equation for the dissociation process is

       NaOH _{(s) } ---> Na^{+}_{(aq)} + OH^{-} _{(aq)}

       The specific heat capacity of the mixture is c_p = 4.18 J g^{-1} C^{-1}

       

The combined mass of the solution is

         M_t = 13 + 250

         M_t = 263.9 \ g

The mass of the surround here is the mass of the coffee-cup calorimeter and this contain the mixture ( water and the NaOH ) so the mass of the surrounding is  

  M_t = 263.9 \ g

       

3 0
3 years ago
Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79
BARSIC [14]

Answer :  The mole fraction and partial pressure of CH_4,CO_2 and He gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of CH_4 = 1.79 mole

Moles of CO_2 = 1.20 mole

Moles of He = 3.71 mole

Now we have to calculate the mole fraction of CH_4,CO_2 and He gases.

\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267

and,

\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179

and,

\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554

Thus, the mole fraction of CH_4,CO_2 and He gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of CH_4,CO_2 and He gases.

According to the Raoult's law,

p_i=X_i\times p_T

where,

p_i = partial pressure of gas

p_T = total pressure of gas  = 5.78 atm

X_i = mole fraction of gas

p_{CH_4}=X_{CH_4}\times p_T

p_{CH_4}=0.267\times 5.78atm=1.54atm

and,

p_{CO_2}=X_{CO_2}\times p_T

p_{CO_2}=0.179\times 5.78atm=1.03atm

and,

p_{He}=X_{He}\times p_T

p_{He}=0.554\times 5.78atm=3.20atm

Thus, the partial pressure of CH_4,CO_2 and He gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0
3 years ago
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