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nata0808 [166]
3 years ago
11

Rb2co3(aq)+fec2h302(aq)-->

Chemistry
1 answer:
nata0808 [166]3 years ago
4 0

Answer:

Rb2CO3(aq)+Fe(C2H3O2)2(aq)--> 2Rb(C2H3O2)(aq) + FeCO3(s)

Explanation:

The reaction shown in the answer is the reaction of rubidium carbonate and iron II acetate. Rubidium is far more reducing than Fe II hence it can displace Fe II from its salt as shown.

The reducing property of metals depends on the value of their individual electrode potential values. For rubidium, its standard reduction potential is -2.98 V while that of Fe II is -0.44V. Hence rubidium can displace Fe II from its salt as shown above.

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Consider the following reaction:
Kitty [74]

Answer : The value of equilibrium constant (Kc) is, 0.0154

Explanation :

The given chemical reaction is:

                        SO_2Cl_2(g)\rightarrow SO_2(g)+Cl_2(g)

Initial conc.    2.4\times 10^{-2}          0             0

At eqm.          (2.4\times 10^{-2}-x)   x              x

As we are given:

Concentration of Cl_2 at equilibrium = 1.3\times 10^{-2}M

That means,

x=1.3\times 10^{-2}M

The expression for equilibrium constant is:

K_c=\frac{[SO_2][Cl_2]}{[SO_2Cl_2]}

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

K_c=\frac{(x)\times (x)}{2.4\times 10^{-2}-x}

K_c=\frac{(1.3\times 10^{-2})\times (1.3\times 10^{-2})}{2.4\times 10^{-2}-1.3\times 10^{-2}}

K_c=0.0154

Thus, the value of equilibrium constant (Kc) is, 0.0154

4 0
3 years ago
Making methanol the element hydrogen is not abundant in nature, but it is a useful reagent in, for example, the potential synthe
stealth61 [152]
ΔG⁰ = ΔH⁰ - TΔS
ΔH⁰ = Hf,(CH₃OH) - Hf,(CO) = -238.7 + 110.5 = -128.2 kJ/mol
ΔS = S(CH₃OH) - S(CO) - 2S(H₂) = 126.8 - 197.7 - 2 x 130.6 = -332.1 J/mol.K
So 
ΔG⁰ = - 128200 + 332.1 T
For the reaction to be spontaneous:
ΔG⁰ < 0
So: -128200 + 332.1 T < 0
332.1 T < 128200
T < 386.028 K
5 0
3 years ago
Am I correct !<br>Please tell me.<br>If not then pls correct !<br>Thanks​
scZoUnD [109]

Cu + 2H2SO4  ⟶ CuSO4 + SO2 + 2H20

In left hand side of the equation.

Cu = 1 atom

H = 4 atoms

S = 2 atoms

O = 8 atoms

In right hand side of the equation.

Cu = 1 atom

S = 2 atoms

0 = 8 atoms

H = 4 atoms

• All the atoms are balanced in the left and right side of the equation and it satisfies the law of conservation of mass.

• Equation is balanced and correct.

*See the attachment .

3 0
2 years ago
How can we tell the distance sediments traveled due to erosion ?​
meriva

Answer:

more rounded the grains are the more they have been moved around

Explanation:

Generally – the more rounded the grains are the more they have been moved around (i.e. the longer the length of time or distance they have moved). Angular grains cannot have travelled far

geolsoc.org.uk

4 0
2 years ago
Calculate the ph of a solution containing 0.0451 m potassium hydrogen tartrate and 0.028 m dipotassium tartrate. The ka values f
Marina86 [1]

Given buffer:

potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )

[KHC4H4O6] = 0.0451 M

[K2C4H4O6] = 0.028 M

Ka1 = 9.2 *10^-4

Ka2 = 4.31*10^-5

Based on Henderson-Hasselbalch equation;

pH = pKa + log [conjugate base]/[acid]

where pka = -logKa

In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2

pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]

     = -log (4.31*10^-5) + log [0.0451]/[0.028]

pH = 4.15



4 0
3 years ago
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