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3 years ago

5

The first periodic table were arranged elements in order of

1 answer:

FrozenT [24]3 years ago

3 0

Answer:

in order of atomic mass

Explanation:

apex

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Which would result in radioactive decay? Check all that apply

Gekata [30.6K]

6+2=115 and its good it took test

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4 years ago

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Choose the products created during nuclear fusion. Check all that apply.

WARRIOR [948]

Answer

a) Energy

b) Helium Gas

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3 years ago

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What are the concentrations of A , A, B , B, and C C at equilibrium if, at the beginning of the reaction, their concentrations a

Elenna [48]

The question is incomplete, here is the complete question:

A reaction

$A+B\rightleftharpoons C$

has a standard free-energy change of -4.88 kJ/mol at 25°C

What are the concentrations of A, B, and C at equilibrium if at the beginning of the reaction their concentrations are 0.30 M, 0.40 M and 0 M respectively?

<u>Answer:</u> The equilibrium concentrations of A, B and C are 0.117 M, 0.217 M, 0.183 M respectively.

<u>Explanation:</u>

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K_c$

where,

$\Delta G^o$ = Standard Gibbs free energy = -4.88 kJ/mol = -4880 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

$K_c$ = equilibrium constant of the reaction

Putting values in above equation, we get:

$-4880J/mol=-(8.3145J/Kmol)\times 298K\times \ln K_c\\\\K_c=7.17$

We are given:

Initial concentration of A = 0.30 M

Initial concentration of B = 0.40 M

Initial concentration of C = 0 M

The chemical reaction follows:

$A+B\rightleftharpoons C$

<u>Initial:</u> 0.30 0.40 0

<u>At eqllm:</u> 0.30-x 0.40-x x

The expression of equilibrium constant for the above reaction follows:

$K_c=\frac{[C]}{[A][B]}$

We are given:

$K_c=7.17$

Putting values in above equation, we get:

$7.17=\frac{x}{(0.30-x)\times (0.40-x)}\\\\x=0.183,0.657$

Neglecting the value of x = 0.657, because change cannot be greater than the initial concentration

So, equilibrium concentration of A = $(0.30-x)=(0.30-0.183)=0.117M$

Equilibrium concentration of B = $(0.40-x)=(0.40-0.183)=0.217M$

Equilibrium concentration of C = $x=0.183M$

Hence, the equilibrium concentrations of A, B and C are 0.117 M, 0.217 M, 0.183 M respectively.

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3 years ago

Predict (to 3 sig figs) the isochoric specific heat of gaseous SF6 (molar mass 146.06 g/mol) at 1200 K, assuming that at this te

bazaltina [42]

Explanation:

As it is known that $SF_{6}$ molecule is a non-linear molecule. Therefore, its isochoric heat capacity will be as follows.

$C_{v} = \frac{3}{2}R + \frac{3}{2}R + (3 \times 7 - 6)R$

= $\frac{3}{2}R + \frac{3}{2}R + 15R$

= 18 R

Also, $C_{V} = M \times C_{v}$

where, $C_{V}$ = molar heat capacity

M = molecular mass

$C_{v}$ = specific heat

Hence, calculate the value of $C_{v}$ as follows.

$C_{V} = M \times C_{v}$

$8 \times 8.314 \times 10^{7} = 146.06 \times C_{v}$

$C_{v} = 10.2 \times 10^{6} erg. K^{-1}. gm^{-1}$

This means that value of isochoric specific heat is $10.2 \times 10^{6} erg. K^{-1}. gm^{-1}$ .

Yes, we have to assume ideal gas behavior because for ideal gas:

dU = $nC_{v}dT$

Whereas for real gases " $\frac{an^{2}}{V^{2}}$ " has to be added here.

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3 years ago

Aldenhydes may oxidize to form

bonufazy [111]

Aldehydes oxidise to form carboxylic acids

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3 years ago