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3 years ago

Odor extensive or intensive

Chemistry

1 answer:

slava [35]3 years ago

6 0

Answer:

Odour is intensive

Explanation:

intensive properties are Independent of the amount of matter

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A monoprotic weak acid, HA , dissociates in water according to the reaction:

Ratling [72]

<u>Answer:</u> The $pKa$ of the acid is 6.09

<u>Explanation:</u>

For the given chemical reaction:

$HA(aq.)\rightleftharpoons H^+(aq.)+A^-(aq.)$

The expression of equilibrium constant [tex[(K_a)[/tex] for the above equation follows:

$K_a=\frac{[H^+][A^-]}{[HA]}$

We are given:

$[HA]_{eq}=0.200M$

$[H^+]_{eq}=4.00\times 10^{-4}M$

$[A^-]_{eq}=4.00\times 10^{-4}M$

Putting values in above expression, we get:

$K_a=\frac{(4.00\times 10^{-4})\times (4.00\times 10^{-4}}{0.200}\\\\K_a=8.0\times 10^{-7}0$

p-function is defined as the negative logarithm of any concentration.

$pKa=-\log(K_a)$

So,

$pKa=-\log(8.0\times 10^{-7})\\\\pKa=6.09$

Hence, the $pKa$ of the acid is 6.09

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3 years ago

Sterling silver is a mixture of the metals silver and copper.It is used to make jewellery.sometimes sterling silver causes green

slavikrds [6]

The composition of 925 Sterling Silver lends itself to the occasional green discoloration because of the presence of copper

8 0

3 years ago

Which scientist bombarded gold sheets with alpha particles to look at the atom?

Alekssandra [29.7K]

the correct scientist is rutherford

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3 years ago

A scientist measures the standard enthalpy change for the following reaction to be 67.9 kJ:

hodyreva [135]

Answer:

-252.5 kJ/mol = ΔH H2O(g)

Explanation:

ΔH Fe2O3 = -825.5kJ/mol

ΔH H2 = 0kJ/mol

ΔH Fe = 0kJ/mol

Based on Hess's law, ΔH of a reaction is the sum of ΔH of products - ΔH of reactants. For the reaction:

Fe2O3(s) + 3 H2(g) →2Fe(s) + 3 H2O(g)

ΔHr = 67.9kJ/mol = 3*ΔH H2O + 2*ΔHFe - (ΔH Fe2O3 + 3*Δ H2)

67.9kJ/mol = 3*ΔH H2O + 2*0kJ/mol - (ΔH -825.5kJ/mol + 3*Δ H2)

67.9 = 3*ΔH H2O(g) + 825.5kJ/mol

-757.6kJ/mol = 3*ΔH H2O(g)

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3 years ago

Complete combustion of 8.10 g of a hydrocarbon produced 25.9 g of CO2 and 9.27 g of H2O. What is the empirical formula for the h

balu736 [363]

CxHy + O2 --> x CO2 + y/2 H2O

Find the moles of CO2 : 18.9g / 44 g/mol = .430 mol CO2 = .430 mol of C in compound

Find the moles of H2O: 5.79g / 18 g/mol = .322 mol H2O = .166 mol of H in compound

Find the mass of C and H in the compound:

.430mol x 12 = 5.16 g C

.166mol x 1g = .166g H

When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.

Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).

In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).

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4 years ago