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sukhopar [10]
3 years ago
15

Define surface tension

Chemistry
1 answer:
Alisiya [41]3 years ago
6 0

Answer:

the tension of the surface film of a liquid caused by the attraction of the particles in the surface layer by the bulk of the liquid, which tends to minimize surface area.

Explanation:

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What is the pOH of 0.50 molar H3BO3?
Crazy boy [7]

<u>Answer:</u>

<em>A. 10.25</em>

<em></em>

<u>Explanation:</u>

Pkb =4.77

So pka = 14 - pka = 9.23

Ka =10^{-pka}

H_3 BO_3 (aq)+ H_2 O(l) H_2 BO_3^- (aq)+H_3 O^+ (aq)

Initial                0.50M                                 0                                 0

Change                  -x                                 +x                               +x

Equilibrium    0.50M-x                               +x                               +x

Ka =\frac {((x)(x))}{(0.50M-x)}

5.88\times10^{-10}= \frac {x^2}{(0.50M-x)}

(-x is neglected) so we get

5.88\times10^{-10}\times0.50=x^2\\\\x^2=2.94\times10^{-10}

x=\sqrt{x^2}=1.72\times10^{-5} M=H^3 O^{+}

pH=-log[H^3 O^+]\\\\pH=-log[1.72\times10^{-5}]\\\\pH=4.76

pOH = 14 - pH

= 14 - 4.76

pOH = 9.24 is the answer

Option A - 10.25 is the answer which is close to 9.24

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Một đơn vị cacbon có khối lượng bằng 0.16605.10²³(g) khối lượng tính bằng gam của nguyên tử Kali (K=39) là:?
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What is the difference between a solution and a heterogeneous mixture? What are some examples?
liubo4ka [24]

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8 0
3 years ago
For the first order process: AB The half-life of A is 62.1 seconds. If a sample of A initially has 250.0 g, what mass (in g) of
vodomira [7]

<u>Answer:</u> The mass of sample A after given time is 99.05 g.

<u>Explanation:</u>

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

We are given:

t_{1/2}=62.1s

Putting values in above equation, we get:

k=\frac{0.693}{62.1}=0.011s^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant = 0.011s^{-1}

t = time taken for decay process = 84.2 s

[A_o] = initial amount of the reactant = 250 g

[A] = amount left after decay process =  ?

Putting values in above equation, we get:

0.011s^{-1}=\frac{2.303}{84.2s}\log\frac{250}{[A]}

[A]=99.05g

Hence, the mass of sample A after given time is 99.05 g.

8 0
3 years ago
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