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sergey [27]
3 years ago
7

What must be true for a substance to be able to dissolve in water?

Chemistry
1 answer:
zvonat [6]3 years ago
3 0

In order for a solute to dissolve in a solvent, the attractive forces between solute particles and the solvent particles must be stronger than the attractive forces between solute-solute and solvent-solvent particles. This is important so that the solute will remain in solution.

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Calculate the molarity and normality of 5.7 g of Ca(OH)2 in 450 mL of solution.
Kisachek [45]
  The  molarity   and  normality  of  5.7 g  of  Ca(OH)2   in  450ml  0f    solution  is  calculated  as  follows

molarity  =  moles/volume in  liters
moles  =mass/molar  mass
=  5.7g/74g/mol  =  0.077moles
molarity =  0.077/450  x1000= 0.17M

Normality =  equivalent  point  x molarity
equivalent  point  of Ca(OH)2  is  2   since  it has  two  Hydrogen  atom

normality  is therefore =  0.17  x2 = 0.34 N
7 0
3 years ago
According to the Periodic Table of Elements, Calcium has:
lana [24]

Answer:

20 protons and 20 electrons

Explanation:

3 0
2 years ago
Select the correct answer.
erastova [34]

Scientific law

Explanation:

A scientific law is a term that refers to a fact that naturally occurs in the universe.

Scientific laws are natural laws and they hold true.

  • Scientific law is a well tested idea that holds through for a range of natural events.
  • Scientific theories are explanations of phenomenons drawn from series of tests.

Learn more:

experiment brainly.com/question/5096428

#learnwithBrainly

3 0
3 years ago
In your own words list the rules for naming hydrocarbons and substituted hydrocarbons. Be detailed in listing all rules includin
Irina18 [472]

Answer :

The basic rules for naming of hydrocarbons are :

First select the longest possible carbon chain.

The longest possible carbon chain should include the carbons of double or triple bonds.

The naming of alkane is done by adding the suffix -ane, alkene by adding the suffix -ene, alkyne by adding the suffix -yne.

The numbering is done in such a way that first carbon of double or triple bond gets the lowest number.

The carbon atoms of the double or triple bond get the preference over the other substituents present in the parent chain.

If two or more similar alkyl groups are present in a compound, the prefixes di-, tri-, tetra- and so on are used to specify the number of times of the alkyl groups in the chain.

8 0
3 years ago
Consider the following reaction: CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at25 ∘C under these conditions: PCO2PCC
Salsk061 [2.6K]

<u>Answer:</u> The \Delta G for the reaction is 54.6 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

  • To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

  • The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.760atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.180atm

Putting values in above equation, we get:

K_p=\frac{(0.760)^2}{0.140\times 0.180}\\\\K_p=22.92

  • To calculate the Gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 22.92

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(22.92))\\\\\Delta G=54659.78J/mol=54.6kJ/mol

Hence, the \Delta G for the reaction is 54.6 kJ/mol

7 0
2 years ago
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