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Maksim231197 [3]
3 years ago
5

Three elements that are similar to sodium​

Chemistry
1 answer:
poizon [28]3 years ago
5 0

Answer:The ones "most" similar would be the ones closest in mass as well. Those would be Lithium (Li) and Potassium (K)

Explanation:

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Why is Cesium more reactive than Lithium??? Please help
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Explanation: Please mark me as brainliest and i hope that helped!

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Humans face a problem because once petroleum reserves begin to run out, there won't be enough fuel to power all the automobiles
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Once we run out of fuel/petroleum/oil we can use hybrid automobiles using solar power or some other type of energy likes these new hybrid cars that can run with power like a battery on a controller and recharge them.
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The number of neutrons in one atom of mercury with mass number 204 is:
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Number of neutrons is 121.
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3 years ago
Carbon disulfide burns in oxygen to yield car- bon dioxide and sulfur dioxide according to the chemical equation cs2(l) 3 o2(g)
vampirchik [111]

Answer: Oxygen is the limiting reagent.

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As can be seen from the given balanced equation:

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Thus O_2 is the limiting reagent as it limits the formation of products. (0.91-0.51)= 0.40 moles of CS_2 will remain as such and thus  CS_2 is an excess reagent.

6 0
3 years ago
A particular reactant decomposes with a half‑life of 109 s when its initial concentration is 0.280 M. The same reactant decompos
Sophie [7]

Answer:

The order of reaction is 2.

Rate constant is 0.0328 (M s)⁻¹

Explanation:

The rate of a reaction is inversely proportional to the time taken for the reaction.

As we are decreasing the concentration of the reactant the half life is increasing.

a) For zero order reaction: the half life is directly proportional to initial concentration of reactant

b) for first order reaction: the half life is independent of the initial concentration.

c) higher order reaction: The relation between half life and rate of reaction is:

Rate = \frac{1}{k[A_{0}]^{(n-1)}}

Half life =K\frac{1}{[A_{0}]^{(n-1)} }

\frac{(halflife_{1})}{(halflife_{2})}=\frac{[A_{2}]^{(n-1)}}{[A_{1}]^{(n-1)} }

where n = order of reaction

Putting values

\frac{109}{231}=\frac{[0.132]^{(n-1)}}{[0.280]^{(n-1)}}

0.472=(0.472)^{(n-1)}

Hence n = 2

halflife=\frac{1}{k[A_{0}]}

Putting values

231=\frac{1}{K(0.132)}

K = 0.0328

4 0
3 years ago
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