Difference between carbon-11 and boron 11

Use the periodic table to select the element that best fits each of the following descriptions. which element below has properti

xenn [34]

Answer: Boron is the element which has properties of both metals and nonmetals.

Explanation:

Metals are defined as the elements which loose electrons to attain stable electronic configuration. They attain positive charge and form cation. Example: Zinc (Zn), Aluminium (Al) , copper (Cu)

Non-metals are defined as the elements which gain electrons to attain stable electronic configuration. They attain negative charge and form anion. Example: Chlorine (Cl) , Sulphur (S)

Metalloids are defined as the elements which show properties of both metals and non-metals. There are 7 metalloids in the periodic table. They are Boron (B) , Silicon (Si) , Germanium (Ge) , Arsenic (As) , Antimony (Sb), Tellurium (Te) and Polonium (Po).

Thus boron is the element which has properties of both metals and nonmetals.

7 0

3 years ago

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Predict what would be formed

Anarel [89]

Answer:

No

Explanation:

T6tgbv. 55678 r4fyx a

8 0

3 years ago

nswer the following questions relating to HCl, CH3Cl, and CH3Br.HCl(g), can be prepared by the reaction of concentrated H2SO4(ag

alexdok [17]

Answer:

It is an example of double displacement reaction.

4.8 g of NaCl is needed to react.

Explanation:

Balanced reaction: $H_{2}SO_{4}(aq.)+2NaCl(s)\rightarrow 2HCl(g)+Na_{2}SO_{4}(aq.)$

Here, oxidation states of H, S, O, Na and Cl do not change during reaction. Hence it is not a redox reaction.

In this reaction, cations and anions of the reactants interchange their partners during reaction. Hence, it is an example of double displacement reaction.

As $H_{2}SO_{4}(aq.)$ remain in excess amount therefore NaCl (s) is the limiting reagent. Hence production of HCl entirely depends on amount of NaCl used.

Molar mass of HCl = 36.46 g/mol

So, 3.0 g of HCl = $\frac{3.0}{36.46}$ mol of HCl = 0.082 mol of HCl

According to balanced equation-

2 moles of HCl are produced from 2 moles of NaCl

So, 0.082 moles of HCl are produced from 0.082 moles of NaCl

Molar mass of NaCl = 58.44 g/mol

So, mass of 0.082 moles of NaCl = $(0.082\times 58.44)$ g = 4.8 g

Hence 4.8 g of NaCl is needed to react.

4 0

3 years ago

Determine the number of molecules in 0.0500 mole of boron tribromide (show work)

OleMash [197]

Answer:

There are 0.5 moles

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number of particles.

To calculate the moles, we use the equation:

Thus there are 0.5 moles

Explanation:

7 0

2 years ago

What is the maximum number of grams of N-acetyl-p-toluidine can be prepared from 70. milliliters of 0.167 M p-toluidine hydrochl

Genrish500 [490]

Answer: The maximum amount of N-acetyl-p-toluidine that can be prepared is 1.7 grams.

Explanation:

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

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of p-toluidine hydrochloride solution = 0.167 M

Volume of solution = 70. mL

Putting values in above equation, we get:

$0.167M=\frac{\text{Moles of p-toluidine hydrochloride}\times 1000}{70}\\\\\text{Moles of p-toluidine hydrochloride}=\frac{0.167\times 70}{1000}=0.0117mol$

The chemical equation for the reaction of p-toluidine hydrochloride and acetic anhydride follows:

$\text{p-toluidine hydrochloride}+\text{Acetic anhydride}\rightarrow \text{N-acetyl-p-toluidine}$

By Stoichiometry of the reaction:

1 mole of p-toluidine hydrochloride produces 1 mole of N-acetyl-p-toluidine

So, 0.0117 moles of p-toluidine hydrochloride will produce = $\frac{1}{1}\times 0.0117=0.0117mol$ of N-acetyl-p-toluidine

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of N-acetyl-p-toluidine = 149.2 g/mol

Moles of N-acetyl-p-toluidine = 0.0117 moles

Putting values in equation 1, we get:

$0.0117mol=\frac{\text{Mass of N-acetyl-p-toluidine}}{149.2g/mol}\\\\\text{Mass of N-acetyl-p-toluidine}=(0.0117g/mol\times 149.2)=1.7g$

Hence, the maximum amount of N-acetyl-p-toluidine that can be prepared is 1.7 grams.

6 0

3 years ago