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mr Goodwill [35]
4 years ago
11

Describe an object that is transparent. What quality makes it transparent?

Chemistry
1 answer:
lapo4ka [179]4 years ago
4 0

Answer: Water

Explanation: Water is an example of transparent because light can pass through it. Air is also an example.

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A. (i think)

Explanation:

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3 years ago
Can formaldehyde molecules dissolve well in water?
Mice21 [21]

Answer:

Formaldehyde is highly soluble in water.

Explanation:

Formaldehyde is the simplest of the aldehydes (R−CHO) it is an organic compound which occurs naturally with the formula H−CHO. . The IUPAC name of formaldehyde is methanal.

Formaldehyde  at room temperature is a colorless gas and the liquid is invisible or like white-water. Formaldehyde has an irritating and very pungent like odour. Formaldehyde is very soluble in water.

Now according to Henry's constant law (3 × 10-5 kPa·m3/mol) Formaldehyde will be very unlikely to volatilize from water.and hence, Formaldehyde is considered to be highly soluble in water.

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4 years ago
Whats does the elements Li (lithium),B (boron, and Ne neon) have in common?
nikdorinn [45]
They are in the same period (horizontal)
8 0
3 years ago
The decomposition of \rm XY is second order in \rm XY and has a rate constant of6.96Ã10â3M^{-1} \cdot s^{-1} at a certain temper
LUCKY_DIMON [66]

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

t_{1/2}=\frac{1}{k\times a_0}

k = rate constant =?

a_0 = initial concentration

a = concentration left after time t

Integrated rate law for second order kinetics is given by:

\frac{1}{a}=kt+\frac{1}{a_0}

a) Initial concentration of XY = a_o=0.100 M

Rate constant of the reaction = k = 6.96\times 10^{-3} M^{-1} s^{-1}

Half life of the reaction is:

t_{1/2}=\frac{1}{6.96\times 10^{-3} M^{-1} s^{-1}\times 0.100 M}

=1,436.78 s

1,436.78 seconds is the half-life for this reaction.

b) Initial concentration of XY = 0.100 M

Final concentration after time t = 12.5% of 0.100 M = 0.0125 M

\frac{1}{0.0125 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.100M}

Solving for t;

t = 10,057.47 seconds

In 10,057.47 seconds the concentration of XY will become 12.5% of its initial concentration.

c) Initial concentration of XY = 0.200 M

Final concentration after time t = 12.5% of 0.200 M = 0.025 M

\frac{1}{0.025 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}

Solving for t;

t = 5,028.73 seconds

In 5,028.73 seconds the concentration of XY will become 12.5% of its initial concentration.

d) Initial concentration of XY = 0.160 M

Final concentration after time t = 6.20\times 10^{-2} M

\frac{1}{6.20\times 10^{-2} M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}

Solving for t;

t = 1,419.40 seconds

In 1,419.40 seconds the concentration of XY will become 6.20\times 10^{-2} M.

e)  Initial concentration of SO_2Cl_2= 0.050 M

Final concentration after time t = x

t = 55.0 s

\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 55.0 s+\frac{1}{0.050 M}

Solving for x;

x = 0.04906 M

The concentration after 55.0 seconds is 0.04906 M.

f) Initial concentration of XY= 0.050 M

Final concentration after time t = x

t = 500 s

\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 500 s+\frac{1}{0.050 M}

Solving for x;

x = 0.04259 M

The concentration after 500 seconds is 0.0.04259 M.

7 0
3 years ago
Please someone help me!
svp [43]

sorry needed the points

3 0
3 years ago
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