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Georgia [21]
4 years ago
15

Alkali metals are found in many common substances but are rarely found in pure form. Explain why this is in the case

Chemistry
1 answer:
Evgesh-ka [11]4 years ago
6 0

Alkali metals are rarely found in pure form because they are so reactive.

Alkali metals are in Group 1 of the Periodic Table.

They need to lose only one valence electron to reach a filled valence shell

Thus, they readily react wirh water and other chemicals in the earth to form compounds.

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86.30 mL of an HCl solution was required to neutralize 31.75 mL of 0.150 M NaOH. Determine the molarity of the HCI.
shepuryov [24]

Answer:

0.055M

Explanation:

Using the formula as follows:

CaVa = CbVb

Where;

Ca = concentration of acid (M)

Cb = concentration of base (M)

Va = volume of acid (mL)

Vb = volume of base (mL)

According to the information given in this question, Ca (HCl) = ?, Cb (NaOH) = 0.150, Va (HCl) = 86.30mL, Vb (NaOH) = 31.75mL

Using CaVa = CbVb

Ca = CbVb ÷ Va

Ca = (0.150 × 31.75) ÷ 86.30

Ca = 4.7625 ÷ 86.30

Ca = 0.055M

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Which participates in chemical reactions?<br><br> electrons<br> protons<br> neutrons<br> gamma rays
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Given the following reaction: 2N2O5=2N2O4+O2, if the rate of oxygen production is 0.15M/min, determine:
astraxan [27]

Answer:

a. r_{N_2O_5}=-0.075M/min

b. r_{N_2O_4}=0.075M/min

Explanation:

Hello.

In this case, according to the balanced chemical reaction, we can write the law of rate proportions:

\frac{r_{N_2O_5}}{-2} =\frac{r_{N_2O_4}}{2} =\frac{r_{O_2}}{1}

Thus, we proceed as follows:

a. Since the rate of oxygen production is 0.15 M/min, we can make the following setup:

\frac{r_{N_2O_5}}{-2}  =\frac{r_{O_2}}{1}\\\\r_{N_2O_5}=\frac{r_{O_2}}{-2}  =\frac{0.15M/min}{-2}\\\\ r_{N_2O_5}=-0.075M/min

b. Since the rate of oxygen production is 0.15 M/min, we can make the following setup:

\frac{r_{N_2O_4}}{2}  =\frac{r_{O_2}}{1}\\\\r_{N_2O_4}=\frac{r_{O_2}}{2}  =\frac{0.15M/min}{2}\\\\ r_{N_2O_4}=0.075M/min

Best regards!

8 0
3 years ago
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