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4 years ago

How many significant figures are in 0.08260 L? 4 5 6

Chemistry

2 answers:

mars1129 [50]4 years ago

6 0

Answer:

Explanation:

Count the digits after the first non zero number, any number after that (even zero) counts.

Hope that helped!!! k

Anni [7]4 years ago

5 0

Answer:

$\boxed{4}$

Explanation:

$\sf Leading \ zeros \ are \ not \ significant.$

$\sf Trailing \ zeros \ are \ significant.$

$\sf Non\ zeros \ are \ significant.$

$0.08260 \sf \ has \ 4 \ significant \ figures.$

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what do dots represent in a electron dot diagram and what does the amount of dots tell us about each atom

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4 years ago

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.

VladimirAG [237]

Answer:

For 1: The standard Gibbs free energy change of the reaction is 10.60 kJ/mol

For 2: The equilibrium constant for the given reaction at 298 K is $1.386\times 10^{-2}$

For 3: The equilibrium pressure of oxygen gas is 0.0577 atm

Explanation:

For 1:

The equation used to calculate standard Gibbs free energy change of a reaction is:

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

For the given chemical reaction:

$M_2O_3(s)\rightarrow 2M(s)+\frac{3}{2}O_2(g)$

The equation for the standard Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(M(s))})+(\frac{3}{2}\times \Delta G^o_f_{(O_2(g))})]-[(1\times \Delta G^o_f_{(M_2O_3(s))})]$

We are given:

$\Delta G^o_f_{(M_2O_3(s))}=-10.60kJ/mol\\\Delta G^o_f_{(M(s))}=0kJ/mol\\\Delta G^o_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (0))+(\frac{3}{2}\times (0))]-[(1\times (-10.60))]\\\\\Delta G^o_{rxn}=10.60kJ/mol$

Hence, the standard Gibbs free energy change of the reaction is 10.60 kJ/mol

For 2:

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = Standard Gibbs free energy = 10.60 kJ/mol = 10600 J/mol (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 298 K

$K_{eq}$ = equilibrium constant = ?

Putting values in above equation, we get:

$10600J/mol=-(8.314J/Kmol)\times 298K\times \ln (K_{eq})\\\\K_{eq}=1.386\times 10^{-2}$

Hence, the equilibrium constant for the given reaction at 298 K is $1.386\times 10^{-2}$

For 3:

The expression of $K_{eq}$ for above equation follows:

$K_{eq}=p_{O_2}^{3/2}$

The concentration of pure solids and pure liquids are taken as 1 in the expression. That is why, the concentration of metal and metal oxide is taken as 1 in the expression.

Putting values in above expression, we get:

$1.386\times 10^{-2}=p_{O_2}^{3/2}\\\\p_{O_2}=0.0577atm$

Hence, the equilibrium pressure of oxygen gas is 0.0577 atm

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3 years ago

Which is a characteristic of all ions?

jonny [76]

Answer:

They are made of one type of atoms.

Explanation:

Every ion that exists has at least one atom. If we didn't have atoms, we would not have ion because atoms make up ions.

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3 years ago

Question 2 (Multiple Choice Worth 2 points)

faust18 [17]

Answer:

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Explanation:

A can is cylindrical in nature. Using the formula of the volume of the can, we can find this unknown volume.

The volume of cylinder is given as:

Volume of a cylinder = Area x height

Volume of a cylinder = $\pi$ r² x h

Therefore density of the can;

Density = $\frac{mass}{volume}$

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3 years ago

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solniwko [45]

Answer:

The boundary would be a divergent boundary

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3 years ago