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grigory [225]
3 years ago
10

What’s a chemical property

Chemistry
1 answer:
abruzzese [7]3 years ago
6 0

Answer:

A chemical property is any of a material's properties that becomes evident during, or after, a chemical reaction; that is, any quality that can be established only by changing a substance's chemical identity. ... They can also be useful to identify an unknown substance or to separate or purify it from other substances.

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Based on the temperature and rainfall, it is the tropical

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Read 2 more answers
Help asap - chemistry, will mark brainlyest
melisa1 [442]

Answer:

Error percentage = 1.326% , 1.35% ,  5.37%

density = mass / volume

For first sample:

density = 37.22/11.5 = 3.2365 g/ml

then percentage error = 100% -  (3.2365/3.28 ) * 100 = 1.326%

For second sample:

density = 37.21/11.5 = 3.2356 g/ml

then percentage error = 100% - ( 3.2356 / 3.28 ) *100 = 1.35%

For third sample:

density = 37.25/12 = 3.104 g/ml

then percentage error = 100% - ( 3.104 / 3.28 ) *100 = 5.37%

3 0
3 years ago
An article in Science News stated, "Noble gases are snobs." What did the author mean?
eduard

Answer:

Explanation:

Noble gases occupy the last group of the periodic table

They have fully filled valence electron shell

What this means is that they have attained stability and thus do not take part in chemical bonding that usually invloves the transfer or sharing of electrons

Thus, noble gases are snobs because they do not partake in chemical bonding with atoms of other elements or atoms of theirselves

8 0
1 year ago
If carbon dioxide is broken down. what element will it give?​
katrin2010 [14]
Carbon dioxide is CO2...if it’s broken down it will be C and O which are carbon and oxygen
7 0
3 years ago
[Co(H2O)6]2+(aq) + 4Cl-(aq) ⇌ [CoCl4]2-(aq) + 6H2O(l)
pav-90 [236]

<u>Answer:</u>

<u>For A:</u> The expression for K_{eq} is given below.

<u>For B:</u> The value of K_{eq} at 25°C is 0.0185512

<u>For C:</u> The value of K_{eq} at 65°C is 0.2887886

<u>For D:</u> The reaction is endothermic in nature.

<u>Explanation:</u>

  • <u>For A:</u>

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{eq}

For the given chemical reaction:

[Co(H_2O)_6]^{2+}(aq.)+4Cl^-(aq.)\rightleftharpoons [CoCl_4]^{2-}(aq.)+6H_2O(l)

The expression of K_{eq} for above equation without the concentration of liquid water is:

K_{eq}=\frac{[CoCl_4]^{2-}}{[Co(H_2O)_6]^{2+}[Cl^-]^4}      ......(1)

The expression is written above.

  • <u>For B:</u>

We are given:

[CoCl_4]^{2-}=0.0334612M

[Co(H_2O)_6]^{2+}=0.966539M

[Cl^-]=1.86616M

Putting values in equation 1, we get:

K_{eq}=\frac{0.0334612}{0.966539\times 1.86616}=0.0185512

Hence, the value of K_{eq} at 25°C is 0.0185512

  • <u>For C:</u>

We are given:

[CoCl_4]^{2-}=0.234625M

[Co(H_2O)_6]^{2+}=0.765375M

[Cl^-]=1.06150M

Putting values in equation 1, we get:

K_{eq}=\frac{0.234625}{0.765375\times 1.06150}=0.2887886

Hence, the value of K_{eq} at 65°C is 0.2887886

  • <u>For D:</u>

For Endothermic reactions, \Delta H>0, which is positive

For Exothermic reactions, \Delta H, which is negative

To calculate \Delta H of the reaction, we use Van't Hoff's equation, which is:

\ln(\frac{K_{65^oC}}{K_{25^oC}})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{65^oC} = equilibrium constant at 65°C = 0.2887886

K_{25^oC} = equilibrium constant at 25°C = 0.0185512

\Delta H = Enthalpy change of the reaction = ?

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 25^oC=[25+2730]K=298K

T_2 = final temperature = 65^oC=[65+2730]K=338K

Putting values in above equation, we get:

\ln(\frac{0.2887886}{0.0185512})=\frac{\Delta H}{8.314J/mol.K}[\frac{1}{298}-\frac{1}{338}]\\\\\Delta H=57471.26J/mol

As, the calculated value of \Delta H>0. Thus, the reaction is endothermic in nature.

4 0
3 years ago
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