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2 years ago

Can Thiocyanic acid, HSCN, can be a Bronsted-Lowry Acid? How about a Bronsted-Lowry Base?

Chemistry

2 answers:

stealth61 [152]2 years ago

7 0

Answer:

It can be a bronsted Lowry acid but not a bronsted Lowry base

Explanation:

Bronsted Lowry acid are proton donors, and HSCN can donate a proton, but it cannot accept a proton, hence not a bronsted Lowry base

Aleksandr-060686 [28]2 years ago

6 0

Answer:

Explanation:

A Brönsted-Lowry acid is defined as any substance that has the ability to lose, or "donate a proton" [H +].

A Brönsted-Lowry base is a substance capable of gaining or "accepting a proton" [H +].

Then a proton transfer occurs, which requires the presence of a proton donor, that is, an acid and a base that accepts them.

This theory has the disadvantage of leaving out several substances that are acidic and that do not have protons.

Thiocyanic acid is a chemical compound that can be considered, but not a Bronsted Lowry base, giving up the proton and generating the anion [SCN] -

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Butoxors [25]

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2 years ago

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An evacuated 276 mL glass bulb weighs 129.6375 g. Filled with an unknown gas, the bulb weighs 130.0318 g. Calculate the gas dens

baherus [9]

Answer:

density=1.43 g/L

Explanation:

Since the density formula is density = mass / volume, we need to find out the mass of the gas and the volume is that of the container.

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2 years ago

What is the ratio of lactic acid (Ka = 1.37x^10-4) to lactate in a solution with pH =4.29

hram777 [196]

Henderson–Hasselbalch equation is given as,

pH = pKa + log [A⁻] / [HA] -------- (1)

Solution:

Convert Ka into pKa,

pKa = -log Ka

pKa = -log 1.37 × 10⁻⁴

pKa = 3.863

Putting value of pKa and pH in eq.1,

4.29 = 3.863 + log [lactate] / [lactic acid]

Or,

log [lactate] / [lactic acid] = 4.29 - 3.863

log [lactate] / [lactic acid] = 0.427

Taking Anti log,

[lactate] / [lactic acid] = 2.673

Result:

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2 years ago

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Lina20 [59]

Answer:

100.8 °C

Explanation:

The Clausius-clapeyron equation is:

$ln\frac{P_{1} }{P_{2}} =$ -Δ $\frac{H_{vap}}{r} (\frac{1}{T_{2}}-\frac{1}{T_{1}} )$

Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'

Isolating for T2 gives:

$T_{2}=(\frac{1}{T_{1}} -\frac{Rln\frac{P_{2}}{P_{1}} }{Delta H_{vap}}$

(sorry for 'deltaHvap' I can not input symbols into equations)

thus T2=100.8 °C

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GuDViN [60]

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