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3 years ago

Which of the following types of mixtures exhibit the Tyndall effect?

1 answer:

4 0

C) Colloids

Colloids have small non-dissolved particles that flow around in the mixture. These particles do not settle over time. When a light is shined on colloids the scattering characteristic of the Tyndall effect are visable.

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Draw a Lewis structure for [H3O]+. Show all unshared pairs and the formal charges, if any.

Montano1993 [528]

Answer: Formal Charges: Hydrogen = 0 and Oxygen = +1

Unshared Pair of electrons: Hydrogen = 0 and Oxygen = 2

Explanation:

The attachment below shows the Lewis structure and the calculations

3 0

3 years ago

A cube has an edge length of 9 cm .

Angelina_Jolie [31]

Answer:

9^3 i think so like 279

Explanation:

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5 0

3 years ago

The nucleus of an atom ok K-42 contains

horsena [70]

Number of proton K=19

so, 42 - 19 =23

then the answer in 19 protons and 23 neutrons

3 0

3 years ago

Describe the chemical reaction based on the

BabaBlast [244]

Answer:

Ammonia gas reacts with oxygen gas.

Nitric oxide gas and liquid water are produced.

Platinum is used as a catalyst.

The equation is unbalanced because the number of hydrogen atoms is not the same on both sides of the equation.

Explanation:

3 0

3 years ago

Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s

bazaltina [42]

Answer: The time taken by the reaction is 84.5 seconds

Explanation:

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

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

where,

$t_{1/2}$ = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{9}=0.077s^{-1}$

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

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$ ......(1)

where,

k = rate constant = $0.077s^{-1}$

t = time taken for decay process = 50.7 sec

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 0.0741 M

Putting values in equation 1, we get:

$0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}$

$[A_o]=3.67M$

Now, calculating the time taken by using equation 1:

$[A]=0.0055M$

$k=0.077s^{-1}$

$[A_o]=3.67M$

Putting values in equation 1, we get:

$0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s$

Hence, the time taken by the reaction is 84.5 seconds

6 0

3 years ago