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neonofarm [45]
3 years ago
10

The total amount of energy that remains in the system always ______ even in an ideal system with no friction or air resistance.

Chemistry
1 answer:
Marizza181 [45]3 years ago
3 0
The amount of energy stays the same
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Firdavs [7]
Having a variety is always good you never want to have too much of one thing. It helps with the food chain its basically survival of the fittest 
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Does CO2 contain molecules?
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5 0
3 years ago
An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 an
Westkost [7]

Answer:

7.3

Explanation:

By Henderson Hasselbalch equation we can calculate the pH or the pOH of a solution by its pKa. Remember that pH = -log[H^{+}], and pKa = -logKa. Ka is the equilibrium constant of the acid.

Henderson Hasselbalch equation :

pH = pKa - log \frac{[HA]}{[A^{-}]}

Where [HA] is the concentration of the acid, and [A^{-}] is the concentration of the anion which forms the acid.

So, acid X, has two ionic forms, the carboxyl group and the other one. First, we have 0.1 mol/L of the acid, in 100 mL, so the number of moles of X

n1 = (0.1 mol/L)x(0.1 L) = 0.01 mol

When it dissociates, it forms 0.005 mol of the carboxyl group and 0.005 mol of the other group. Assuming same  stoichiometry.

Adding NaOH, with 0.1 mol/L and 75 mL, the number of moles of OH^- will be

n2 = (0.1 mol/L)x(0.075 L) = 0.0075 mol

So, the 0.0075 mol of OH^- reacts with 0.005 mol of carboxyl, remaining 0.0025 mol of OH^-, which will react with the 0.005 mol of the other group. So, it will remain 0.0025 mol of the other group.

The final volume of the solution will be 175 mL, but both concentrations (the acid form and ionic form) have the same volume, so we can use the number of mol in the equation.

Note that, the number of moles of the acid form is still 0.01 mol because it doesn't react!

So,

6.72 = pKa - log \frac{0.01}{0.0025}

6.72 = pKa - log 4

pKa - log4 = 6.72

pKa = 6.72 + log4

pKa = 6.72 + 0.6

pKa = 7.3

8 0
3 years ago
Calcule la densidad del hidrógeno H2 en g/L a 327 mm Hg y 48ºc
tankabanditka [31]
<h3>The density of H₂ = 0.033 g/L</h3><h3>Further explanation</h3>

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm , N/m²

V = volume, liter  

n = number of moles  

R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m², v= m³)

T = temperature, Kelvin  

n = N / No  

n = mole  

No = Avogadro number (6.02.10²³)  

n = m / MW

m = mass  

MW = molecular weight

For density , can be formulated :

\tt \rho=\dfrac{P\times MW}{R\times T}

P = 327 mmHg = 0,430263 atm

R = 0.082 L.atm / mol K

T = 48 ºC = 321.15 K

MW of H₂ =  2.015 g/mol

The density :

\rho=\dfrac{0,430263\times 2.015 }{0.082\times 321.15}\\\\\rho=0.033~g/L

4 0
2 years ago
When strongly electronegative atoms, like fluorine, bond to atoms with a lower electronegativity, like hydrogen, what's the resu
olganol [36]
When highly electronegative element like oxygen is directly attached to less electronegative element like hydrogen the electrons from less electronegative elements are attracted toward the highly electronegative element, making the less electronegative element deficient in electron density (partial positive) and a partial negative charge on more electronegative element is created. In such situation the intermolecular forces formed are dipole-dipole interactions or hydrogen bond interaction like in HF.
7 0
3 years ago
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