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2 years ago

According to newtons third law forces always occur in equal but____ pairs?

Chemistry

1 answer:

arsen [322]2 years ago

8 0

'Every action has an equal but opposite reaction' They are equal, but opposite.

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What is a system of knowledge and the methods of find that knowledge

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Answer:

A system of knowledge and the methods to find that knowledge is Science.

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2 years ago

Lice are parasitic insects that live on the scalp, eyebrows, and eyelashes of humans

earnstyle [38]

Answer:

Most likely A

Explanation:

Its a more plausible answer

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3 years ago

If you have a negative ion, what happened to the outer shell of the atom?

Nastasia [14]

It will lose them and become stable

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2 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

Now put all the given values in this expression, we get:

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

4 0

3 years ago

You could add solid KCl to the solution to precipitate out AgCl(s). What mass of KCl is needed to precipitate the silver ions fr

padilas [110]

Answer:

0.143 g of KCl.

Explanation:

Equation of the reaction:

AgNO3(aq) + KCl(aq) --> AgCl(s) + KNO3(aq)

Molar concentration = mass/volume

= 0.16 * 0.012

= 0.00192 mol AgNO3.

By stoichiometry, 1 mole of AgNO3 reacts with 1 mole of KCl to form a precipitate.

Number of moles of KCl = 0.00192 mol.

Molar mass of KCl = 39 + 35.5

= 74.5 g/mol

Mass = molar mass * number of moles

= 74.5 * 0.00192

= 0.143 g of KCl.

4 0

3 years ago