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3 years ago

In which atmosphere layer do humans live most of their life

Chemistry

1 answer:

marysya [2.9K]3 years ago

3 0

Troposphere, this is the layer of the atmosphere closest to the earths crust.

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White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white

Alex_Xolod [135]

The pH = 2.41

<h3>Further explanation</h3>

Given

5.0% by mass solution of acetic acid

the density of white vinegar is 1.007 g/cm3

Required

Solution

Molarity of solution :

$\tt M=\dfrac{\%mass\times \rho\times 10}{MW~acetic~acid}\\\\M=\dfrac{5\times 1.007\times 10}{60}\\\\M=0.839$

Ka for acetic acid = 1.8 x 10⁻⁵

[H⁺] for weak acid :

$\tt [H^+]=\sqrt{Ka.M}$

Input the value :

$\tt [H^+]=\sqrt{1.8\times 10^{-5}\times 0.839}\\\\(H^+]=0.00388=3.88\times 10^{-3}\\\\pH=3-log~3.88=2.41$

7 0

3 years ago

{Plz answer} Thank you!

Dafna11 [192]

The Answer to your question would be A

8 0

3 years ago

Read 2 more answers

H4X - pKa1 =1.5

Nataly [62]

Othu['oi;eejlktmkolhkjdfm;hojnor;'mor;'mor;'mor;'mor;'mor;'mhim

8 0

3 years ago

How many atoms of Carbon are found on the PRODUCTS side?

Semenov [28]

Secrets undiscovered is correct, but this chemical formula is not balanced. Double check your question to make sure

5 0

3 years ago

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

7 0

3 years ago