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2 years ago

Define transparent ,translucent ,opaque materials?

2 answers:

5 0

Transparent - when light is in context with transparent materials almost all of the the light is allowed to go through these materials.
Ex: glass

Translucent- materials that do not allow light to pass through easily.
Ex: tissue paper

Opaque- materials that allow no light to go through them .
Ex: wood

patriot [66]2 years ago

3 0

A transparent material is a material that

allows you to see right through it.

For example, glass is usually a transparent material.

A translucent material is a material that allows

you to see light, but it's distorted.

For example, tracing paper is a translucent material.

If a material is opaque, it blocks light from going through them.

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Chlorine forms a number of oxides with the following oxidation numbers: 1, 3, 4, 6, and 7. These compounds are stable and neutra

SashulF [63]

Answer:

$Cl_2O$ , $Cl_2O_3$ , $ClO_2$ , $ClO_3$ , $Cl_2O_7$

Explanation:

Empirical formula of the compound is the simplest ratio of elements present in the compound.

Empirical formula of compounds of chlorine with oxygen is as follows:

Compounds in which oxidation state of Cl is +1

$Cl_2O$

Compounds in which oxidation state of Cl is +3

$Cl_2O_3$

Compounds in which oxidation state of Cl is +4

$ClO_2$

Compounds in which oxidation state of Cl is +6

$ClO_3$

Compounds in which oxidation state of Cl is +7

$Cl_2O_7$

8 0

3 years ago

2NaCl + F2 yields 2NaF + Cl2

bearhunter [10]

Study your experiment setup.<span> In 30 minutes, how will the air temperature in the bottles compare?</span><span> What do you predict will happen to the ice in each bottle?</span>

7 0

2 years ago

I need help with this question please someone tell me the answer for this

Andrej [43]

Aueyeowosy yo this is my answer

5 0

2 years ago

Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp

gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

$ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})$

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in j/mol R Ea is

Ea=35.5*1000 j/mol R

$ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492$

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}