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2 years ago

Why does graphite and diamond sublime?

Chemistry

1 answer:

GREYUIT [131]2 years ago

8 0

Answer:

The intermolecular bonding in the solid is van der Waals dispersion forces. That means that it will have a lower melting point than diamond or graphite where you have giant covalent bonding. In fact C60 sublimes at 800 K (527°C). Diamond and graphite sublime at around 4000 K.

Explanation:

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Ozone is made of oxygen. What is the formula of ozone if 36g occupies 16.8L at STP?

Scilla [17]

The formula of ozone is O3.
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2 years ago

If a bouncing ball has a total energy of 20 J and a kinetic energy of 5 J, the ball’s potential energy is____J.

astra-53 [7]

Answer:If a bouncing ball has a total energy of 20 J and a kinetic energy of 5 J, the ball’s potential energy is 15J.

If the kinetic energy of the ball decreases, then the potential energy will Increase.

Explanation:

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2 years ago

Joyce poured 200 milliliters (mL) of water into a beaker, placed a thermometer in it, and heated it until it started to boil. Sh

earnstyle [38]

Answer:

100°C

Explanation:

The boiling point of water is 100°C and the difference is when will the water start boiling.

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2 years ago

How many grams of NH3 can be prepared from 85.5 grams of N2 and 17.3 grams of H2 ?

Tcecarenko [31]

N2 + 3H2 ---> 2NH3

mass of N2 = 28g
mass of H2 = 2g
mass of NH3 = 17g

according to the reaction:
28g N2----------------- 3*2g H2
85,5g N2-------------------- x
x = 18,32g H2 >>> so, nitrogen is excess

according to the reaction:
2*3g H2---------------------- 2*17g NH3
17,3g H2 ------------------------- x
x = 98,03g NH3

<u>answer: 98,03g of NH3</u>

4 0

2 years ago

A 600.0 mL sample of 0.20 MHF is titrated with 0.10 MNaOH. Determine the pH of the solution after the addition of 600.0 mL of Na

Leona [35]

Answer: pH=12.69

Explanation:

${\text{Moles of HF}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of HF}=0.20M\times 0.6L=0.12 moles$

$HF\rightarrow H^++F^-$

Initial 0.12 0 0

Eqm 0.12-x x x

$K_a=\frac{[H^+][F^-]}{[HF]}$

$3.5\times 10^{-4}=\frac{x^2}{0.12-x}$

(neglecting small value of x in comparison to 0.12)

$x=4.2\times 10^{-5}$

Moles of $H^+=4.2\times 10^{-5}$

$NaOH\rightarrow Na^++OH^-$

${\text{Molesof NaOH}}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of NaOH}}=0.10M\times 0.6L=0.06 moles$

0.06 moles of NaOH will give 0.06 moles of $[OH^-]$

Now $4.2\times 10^{-5}$ moles of $OH^-$ will be neutralized by $4.2\times 10^{-5}$ moles of $H^+$ and $(0.06-4.2\times 10^{-5})=0.059$ moles of $OH^-$ will be left.

Molarity of $OH^-=\frac{0.059moles}{1.2L}=0.049M$

$pOH=-\log[OH^-]=-\log[0.049]=1.31$

pH = 14 - pOH= 14 - 1.31 = 12.69

5 0

2 years ago