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AlladinOne [14]

2 years ago

12

In these chemical compounds, c is for carbon, o is for oxygen, n is for nitrogen, and k is for potassium. Which chemical compoun

d is organic?.

1 answer:

Jet001 [13]2 years ago

8 0

Answer:

carbon because organic compounds are made up of hydrogen and carbon

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Given, 2Ca + O2 -> 2CaO, what is the 2 located in front of calcium called??

EleoNora [17]

Moles used to react

3 0

3 years ago

Given the elements chlorine, iodine, oxygen, bromine, and fluorine, organize by increasing atomic size (atomic radius). Justify

astra-53 [7]

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Explanation:

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2 years ago

Given the equation representing a reversible reaction:

tester [92]

Answer:

- Acetic acid (CH₃COOH) and hydronium ion (H₃O⁺)

Explanation:

Hello,

In this case, based on the acid-base theory which states that acids are known as H⁺ donors, if we consider the direct reaction:

$CH_3COOH(aq) + H_2O \rightarrow CH_3COO^-(aq) + H_3O^+(aq)$

It is clear that the acetic acid is the first H⁺ donor as it losses one H⁺ to turn into the acetate ion. Moreover, if we consider the inverse reaction:

$CH_3COO^-(aq) + H_3O^+(aq)\rightarrow CH_3COOH(aq) + H_2O$

It is also clear that the hydronium ion is the second H⁺ donor as it losses one H⁺ to turn into water.

Best regards.

5 0

2 years ago

Si2Br6 compound name

Anna71 [15]

Answer:

disilicon hexabromide

Explanation:

6 0

2 years ago

A 0.4272 g sample of an element contains 2.241 x 10 ^21 atoms . what is the symbol of the element?

grin007 [14]

Answer:

Likely $\rm In$ (indium.)

Explanation:

Number of atoms: $N = 2.241 \times 10^{21}$ .

Dividing, $N$ , the number of atoms by the Avogadro constant, $N_A \approx 6.023 \times 10^{23} \; \rm mol^{-1}$ , would give the number of moles of atoms in this sample:

$\begin{aligned} n &= \frac{N}{N_{A}} \\ &\approx \frac{2.241 \times 10^{21}}{6.023 \times 10^{23}\; \rm mol^{-1}} \approx 3.72 \times 10^{-3}\; \rm mol \end{aligned}$ .

The mass of that many atom is $m = 0.4272\; \rm g$ . Estimate the average mass of one mole of atoms in this sample:

$\begin{aligned}M &= \frac{m}{n} \\ &\approx \frac{0.4272\; \rm g}{3.72 \times 10^{-3}\; \rm mol} \approx 114.82\; \rm g \cdot mol^{-1}\end{aligned}$ .

The average mass of one mole of atoms of an element ( $114.82\; \rm g \cdot mol^{-1}$ in this example) is numerically equal to the average atomic mass of that element. Refer to a modern periodic table and look for the element with average atomic mass $114.82$ . Indium, $\rm In$ , is the closest match.

5 0

3 years ago