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Converting 53.3g of oxygen to moles will give you???

Marat540 [252]

Answer:

3.33 mol

Explanation:

1 g of oxygen is 0.062502343837894 mol

53.3 * 0.062502343837894

3.3313749265597505 mol

8 0

3 years ago

What’s a black dwarf

Helga [31]

Answer:

you

Explanation:

7 0

4 years ago

Read 2 more answers

Calculate the concentration expressed in percent, if 10 g of NaOH is diluted to 500 ml with water.

Zepler [3.9K]

If 10 g of NaOH is diluted to 500 ml with water then the concentration expressed in percent is 0.5 mol/L .

Calculation ,

Given mass in gram = 10 g

Number of moles = given mass /molar mass = 10 g / 40 g/mol = 0.25mole

Given volume in ml = 500 ml

Given volume in liter = 0.5 L

Putting the value of mass and volume in equation i we get concentration expressed in percent .

C = number of moles ×100/ volume in liter = 0.25mole ×100/ 0.5 L

C = 0.5 mol/L

Concentration of solution in terms of percentage can be expressed in two ways

1) percentage by mass

2) percentage by volume

Hence, for liquid solutions, concentration is expressed in terms of percentage by volume.

To learn more about concentration please click here ,

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5 0

1 year 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

4 years ago

2. Perception:

sineoko [7]

Answer:

D

Explanation:

Someone's perception can be all of these things. the way one person sees something is different than the way someone standing right beside them sees it.

4 0

3 years ago