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aleksandrvk [35]

3 years ago

13

I need help filling out the chart where it says lithium please!!!

1 answer:

vovangra [49]3 years ago

6 0

Use a periodic table, It should help

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Consider the following reaction of calcium hydride (CaH2) with molten sodium metal: CaH2(s) + 2 Na(l) -> 2 NaH(s) + Ca(l) Ide

vodka [1.7K]

Answer:

This is not a redox reaction. None of the species are reduced and none are oxidized

Explanation:

In a redox reaction at least one species must be oxidized and another reduced. You see this by a change in oxidation number. In this question the oxidation numbers are the same before and after the reaction.

Ca 2

H -1

Na +1

I -1

4 0

4 years ago

Nickel and carbon monoxide react to form nickel carbonyl, like this: (s)(g)(g) At a certain temperature, a chemist finds that a

horsena [70]

The question is incomplete, here is the complete question:

Nickel and carbon monoxide react to form nickel carbonyl, like this:

$Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)$

At a certain temperature, a chemist finds that a 2.6 L reaction vessel containing a mixture of nickel, carbon monoxide, and nickel carbonyl at equilibrium has the following composition:

Compound Amount

Ni 12.7 g

CO 1.98 g

$Ni(CO)_4$ 0.597 g

Calculate the value of the equilibrium constant.

<u>Answer:</u> The value of equilibrium constant for the reaction is 2448.1

<u>Explanation:</u>

We are given:

Mass of nickel = 12.7 g

Mass of CO = 1.98 g

Mass of $Ni(CO)_4$ = 0.597 g

Volume of container = 2.6 L

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Given mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

$\text{Equilibrium concentration of nickel}=\frac{12.7}{58.7\times 2.6}=0.083M$

$\text{Equilibrium concentration of CO}=\frac{1.98}{28\times 2.6}=0.0272M$

$\text{Equilibrium concentration of }Ni(CO)_4=\frac{0.597}{170.73\times 2.6}=0.00134M$

For the given chemical reaction:

$Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)$

The expression of equilibrium constant for the reaction:

$K_{eq}=\frac{[Ni(CO)_4]}{[CO]^4}$

Concentrations of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

$K_{eq}=\frac{0.00134}{(0.0272)^4}\\\\K_{eq}=2448.1$

Hence, the value of equilibrium constant for the reaction is 2448.1

4 0

4 years ago

I need this ASAP thank you

son4ous [18]

The first one I god it wrong and he told us the answer

4 0

3 years ago

Which atoms likely to react ? PLS HELPP

Irina18 [472]

Answer:

i can't, it's blocked

Explanation:

8 0

3 years ago

Titration Lab Sheet: Day 2 (Alternate)‼️‼️‼️

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Hey did you ever find the answers to this?

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3 years ago

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