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4 years ago

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Why are atoms of lithium, sodium, and potassium almost neverfound alone in nature

2 answers:

yuradex [85]4 years ago

6 0

<span> The Alkali elements are so reactive that they are </span><span>never found alone in nature</span>

Crazy boy [7]4 years ago

4 0

These three metals- called the alkali metals- have one electron in their outer shell. Since all atoms want to be stable, it try's to lose this electron. They are very reactive, and so will be found in compounds or mixtures, such as sodium chloride (table salt!)

Hope this helps:)

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the reaction of 50 mL of gas with 50 mL of gas via the equation: Cl2(g) + C2H4(g) ➔ C2H4Cl2 (g) will produce a total of ________

spin [16.1K]

Explanation:

The given data is as follows.

50 ml of $Cl_{2}$ , 50 ml of $C_{2}H_{4}$

And, it is known that at STP 1 mole of a gas occupies 22.4 L. Hence, moles present in 50 ml of gas are as follows.

$\frac{50}{22.4 \times 1000}$ (As 1 L = 1000 ml)

= $2.23 \times 10^{-3}$ moles

So, according to the given equation $2.23 \times 10^{-3}$ moles of $Cl_{2}$ reacts with $2.23 \times 10^{-3}$ moles of $C_{2}H_{4}$ .

Hence, moles of $C_{2}H_{4}Cl_{2}$ is equal to the moles of $C_{2}H_{4}$ and $Cl_{2}$ .

Therefore, moles of $C_{2}H_{4}Cl_{2}$ = $2.23 \times 10^{-3}$ moles

1 mole of $C_{2}H_{4}Cl_{2}$ = 22.4 L

$2.23 \times 10^{-3}$ moles = $22.4 \times 2.23 \times 10^{-3} moles$

= 50 ml of product

Thus, we can conclude that 50 ml of products if pressure and temperature are kept constant.

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If 22.99 g of sodium and 35.45 g of chlorine fully react, how much sodium chloride forms?

matrenka [14]

Answer:

The answer to your question is: HCl = 58.44 g

Explanation:

To solve this problem, we must remember the law of conservation of mass that says that, matter cannot be created or destroyed, only changes its form.

That means that, the mass of the reactants must be equal to the mass of the products.

Process

Sodium (Na) + Chlorine (Cl) ⇒ Sodium chloride (HCl)

22.99 g 35.45 g x

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