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8_murik_8 [283]
2 years ago
13

The pOH of a solution is 6.0. Which statement is correct? Use p O H equals negative logarithm StartBracket upper O upper H super

script minus EndBracket. and p H plus P O H equals 14.. The pH of the solution is 20.0. The concentration of OH– ions is 1.0 times 10 to the negative 8 moles per liter.. The concentration of OH– ions is 1.0 times 10 to the 6 moles per liter.. The pH of the solution is 8.0.
Chemistry
2 answers:
castortr0y [4]2 years ago
6 0

Answer:

The pH of the solution is 8.0.

Explanation:

taking the test rn

Natali5045456 [20]2 years ago
4 0

Answer: D.The pH of the solution is 8.0.

Explanation: taking the test!

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A sample of water at 20 degrees c contains what bonds
vladimir2022 [97]
Covalent and hydrogen bonds

3 0
3 years ago
Hydrogen gas was collected by water displacement. what was pressure of the h2 collected if the temperature was 26°c?
Vedmedyk [2.9K]
The ideal gas law may be written as
p= \frac{\rho R T}{M}
where
p = pressure
ρ =density
T = temperature
M = molar mass
R = 8.314 J/(mol-K)

For the given problem,
ρ = 0.09 g/L = 0.09 kg/m³
T = 26°C = 26+273 K = 299 K
M = 1.008 g/mol = 1.008 x 10⁻³ kg/mol

Therefore
p= \frac{(0.09 \, kg/m^{3})*(8.314 \, J/(mol-K))*(299 \, K)}{1.008 \times 10^{-3} \, kg/mol} =2.2195 \times 10^{5} \, Pa

Note that 1 atm = 101325 Pa
Therefore
p = 2.2195 x 10⁵ Pa
   = 221.95 kPa 
   = (2.295 x 10⁵)/101325 atm
   = 2.19 atm

Answer:
2.2195 x 10⁵ Pa (or 221.95 kPa or 2.19 atm)

4 0
3 years ago
Can someone help me please
lesantik [10]
It is an endothermic reaction because the products hAve more heat than the reactions so it was a gain of heat which makes the enthalpy Change positive !
4 0
3 years ago
The activation energy for proline isomerization of a peptide depends on the identity of the preceding residue and obeys Arrheniu
MAVERICK [17]

Answer:

Activation energy of phenylalanine-proline peptide is 66 kJ/mol.

Explanation:

According to Arrhenius equation-     k=Ae^{\frac{-E_{a}}{RT}}    , where k is rate constant, A is pre-exponential factor, E_{a} is activation energy, R is gas constant and T is temperature in kelvin scale.

As A is identical for both peptide therefore-

                                   \frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}

Here \frac{k_{ala-pro}}{k_{phe-pro}}=\frac{0.05}{0.005} , T = 298 K , R = 8.314 J/(mol.K) and E_{a}^{ala-pro}=60kJ/mol

So, \frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}

   \Rightarrow E_{a}^{phe-pro}=65705J/mol=66kJ/mol (rounded off to two significant digit)

So, activation energy of phenylalanine-proline peptide is 66 kJ/mol

7 0
3 years ago
MULTIPLE CHOICE
Flauer [41]

Answer:

its very simple ans we have 2 just multiply256

6 0
3 years ago
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