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3 years ago

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Is elemental phosphorus dull or high luster?

1 answer:

Ugo [173]3 years ago

7 0

Each element<span> can usually be classified as a metal or a non-metal based on their ... They are usually </span>dull<span>and therefore show no metallic </span>luster<span> and they do not reflect ... </span>Dull<span>, Brittle solids; Little or no metallic </span>luster<span>; </span>High<span> ionization energies; </span>High<span> ...</span>

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How would you prepare 2.5 L of a 0.800M solution of KNO3?

statuscvo [17]

First find the mass of <span>solute:

Molar mass KNO</span>₃ = <span>101.1032 g/mol

mass = Molarity * molar mass * volume

mass = 0.800 * 101.1032 * 2.5

mass = 202.2064 g of KNO</span>₃

<span>To prepare 2.5 L (0800 M) of KNO3 solution, must weigh 202.2064 g of salt, dissolve in a Beker, transfer with the help of a funnel of transfer to a volumetric flask, complete with water up to the mark, capping the balloon and finally shake the solution to mix.</span>

hope this helps!

7 0

3 years ago

When comparing the two chair conformations for a monosubstituted cyclohexane ring, which type of substituent shows the greatest

Aleksandr [31]

Answer:

See the explanation

Explanation:

In this case, we have to keep in mind that in the monosubstituted product we only have to replace 1 hydrogen with another group. In this case, we are going to use the methyl group $CH_3$ .

In the axial position, we have a more steric hindrance because we have two hydrogens near to the $CH_3$ group. If we have <u>more steric hindrance</u> the molecule would be <u>more unstable</u>. In the equatorial positions, we don't <u>any interactions</u> because the $CH_3$ group is pointing out. If we don't have <u>any steric hindrance</u> the molecule will be <u>more stable</u>, that's why the molecule will <u>the equatorial position.</u>

See figure 1

I hope it helps!

6 0

3 years ago

In the presence of excess oxygen, methane gas burns in a constant-pressure system to yield carbon dioxide and water:

Marina CMI [18]

Explanation:

in the presence of excess oxygen methane burns to prduce carbon (iv) oxide and water ....this is called complete combustion...

6 0

3 years ago

A comic book villain is holding you at gun point and is making you drink a sample of acid he gives you a beaker with 100 ml of a

miss Akunina [59]

Answer:

b. 10 mL

Explanation:

First we <u>calculate the amount of H⁺ moles in the acid</u>:

pH = -log [H⁺]

[H⁺] = $10^{-pH}$

[H⁺] = 10⁻⁵ = 1x10⁻⁵M

100 mL ⇒ 100 / 1000 = 0.100 L

1x10⁻⁵M * 0.100 L = 1x10⁻⁶ mol H⁺

In order to have a neutral solution we would need the same amount of OH⁻ moles.

We can use the pOH value of the strong base:

pOH = 14 - pH

pOH = 14 - 10 = 4

Then we <u>calculate the molar concentration of the OH⁻ species in the basic solution</u>:

pOH = -log [OH⁻]

[OH⁻] = $10^{-pOH}$ = 1x10⁻⁴ M

If we use 10 mL of the basic solution the number of OH⁻ would be:

10 mL ⇒ 10 / 1000 = 0.010 L

1x10⁻⁴ M * 0.010 L = 1x10⁻⁶ mol OH⁻

It would be equal to the moles of H⁺ so the answer is b.

7 0

3 years ago

Read 2 more answers

In a coffe cup calorimeter, 50.0mL of 0.100M of AgNO3 and 50mL of 0.100M HCl are mixed to yield the following reaction:

Jet001 [13]

Answer:

The enthalpy change of the reaction is -66.88 kJ/mol.

Explanation:

Mass of the solution = m = 100 g

Heat capacity of the solution = c = 4.18 J/g°C

Initial temperature of the solutions before mixing = $T_1=22.60^oC$

Final temperature of the solution after mixing = $T_2=23.40^oC$

Heat gained by the solution due to heat released by reaction between HCl and silver nitrate = Q

$Q=m\times c\times (T_2-T_1)$

$Q=100 g\times 4.18 J/g^oC\times (23.40^oC-22.60^oC)=334.4 J$

Heat released due to reaction = Q' =-Q = -334.4 J

Moles of silver nitrate = n

Molarity of silver nitrate solution = 0.100 M

Volume of the silver nitrate solution = 50.0 mL = 0.050 L ( 1 mL = 0.001 L)

$Moles =Molarity\times Volume (L)$

$n=0.100 M\times 0.050 L=0.005 mol$

Enthalpy change of the reaction = $\Delta H$

$=\Delta H=\frac{-334.4 J}{0.005 mol}=-66,880 J/mol=-66.88 kJ/mol$

1 J = 0.001 kJ

The enthalpy change of the reaction is -66.88 kJ/mol.

7 0

3 years ago