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fenix001 [56]
3 years ago
15

Normal cellular pH has a hydrogen concentration of 3.98 x 10-8 M. what is the pH of a cell

Chemistry
1 answer:
geniusboy [140]3 years ago
6 0
The  Ph of the cell is = 7.4

Ph  = -log  (H^+  concentration)

Ph  =-  log ( 3.98 x10^-8)  =  7.4


this mean the cell is basic  since the PH>7
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Which phase transition, solid to liquid or liquid to gas, results in the largest increase in entropy
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The liquid to gas phase transition results in the largest increase in entropy.

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2 years ago
What is electronic configuration? What is the Law of Constant Composition?
Dovator [93]

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3 0
3 years ago
How many moles of water are in a beaker with 50 mL?
tatiyna

Answer:

Number of moles = 2.8 mol

Explanation:

Given data:

Number of moles of water = ?

Volume of water = 50 mL

Density of water = 1.00 g/cm³

Solution:

1 cm³ =  1 mL

Density = mass/ volume

1.00 g/mL = mass/ 50 mL

Mass = 1.00 g/mL× 50 mL

Mass = 50 g

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Number of moles = mass/molar mass

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5 0
3 years ago
The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3
9966 [12]

Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Where,

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol  

k_2=2.3\times 10^8

T_2=280\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (280 + 273.15) K = 553.15 K  

T_2=553.15\ K  

k_1=4.8\times 10^8  

T_1=376\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (376 + 273.15) K = 649.15 K  

T_1=649.15\ K  

So,  

\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)

E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol

E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol

E_a=22.87\ kJ/mol

<u>The activation energy for this reaction = 23 kJ/mol.</u>

6 0
3 years ago
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