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balu736 [363]
3 years ago
11

Which is the strongest reducing agent in the periodic table?

Chemistry
2 answers:
ycow [4]3 years ago
8 0
"Caesium" is the strongest reducing agent in Periodic table.

Hope this helps!
attashe74 [19]3 years ago
5 0
The strongest reducing agent in the periodic table is Carbon
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A mixture of gaseous CO and H2, called synthesis gas, is used commercially to prepare methanol (CH3OH), a compound considered an
mr_godi [17]

Answer : The value of equilibrium constant (K) is, 424.3

Explanation :  Given,

Concentration of H_2 at equilibrium = 0.067 mol

Concentration of CO at equilibrium = 0.021 mol

Concentration of CH_3OH at equilibrium = 0.040 mol

The given chemical reaction is:

CO+2H_2\rightarrow CH_3OH

The expression for equilibrium constant is:

K_c=\frac{[CH_3OH]}{[CO][H_2]^2}

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

K_c=\frac{(0.040)}{(0.021)\times (0.067)^2}

K_c=424.3

Thus, the value of equilibrium constant (K) is, 424.3

4 0
3 years ago
Why was Becquerel’s experiment a first step in discovering radioactivity?
galina1969 [7]
This was because the experiment showed that a substance could emit radiation even while it was not exposed to light.
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Please help. Thank you guys for all your help and support. <3
miv72 [106K]
It would be “To obey the law of conservation of mass”
7 0
3 years ago
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_H2+O2=_H2O<br> balance help please
erastovalidia [21]
Both are 2
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5 0
3 years ago
How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?
xenn [34]

Answer:

5.225x10^{22}atoms\ I

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I

Best regards!

4 0
3 years ago
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