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dem82 [27]
3 years ago
9

Diamond, graphite, and fullerenes share what property? A. They are all made of carbon (C) bonded to a metal. B. Their shape. C.

They are all made of carbon (C). D. They are all good conductors.
Chemistry
1 answer:
Vika [28.1K]3 years ago
7 0

Answer:

They are all made of carbon

Explanation:

Diamond, graphite, and fullerenes are all made of carbon.

Hope this helps!

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What is the total number of electrons shown in this Lewis structure of carbon dioxide? A. 8 B. 12 C. 16 D. 24
krek1111 [17]
Answer is 16
carbon valence = 4
oxygen valence = 6 but carbon dioxide have 2 oxygen so = 12
12 + 4 = 16
so carbon dioxide have 16 electron
8 0
4 years ago
What determines crystal size in minerals formed by lava or magma?
Burka [1]

Answer:

When magma cools, crystals form because the solution is super-saturated with respect to some minerals. If the magma cools quickly, the crystals do not have much time to form, so they are very small. If the magma cools slowly, then the crystals have enough time to grow and become large.

Explanation:

5 0
3 years ago
How would you describe a food chain to someone who never thought about that idea before
umka21 [38]
A food chain is a sequence going from the producers on the bottom to the consumers to the top that shows what consumer eats what
6 0
3 years ago
Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th
mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

m_{solute} = Given mass of solute (anthracene) = 7.99 g

M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

8 0
3 years ago
If a buffer solution is 0.220 M in a weak acid ( Ka=7.4×10−5) and 0.540 M in its conjugate base, what is the pH?
valkas [14]

Answer: the pH of the solution is 4.52

Explanation:

Consider the weak acid as Ha, it is dissociated as expressed below

HA     H⁺  +  A⁻

the Henderson -Haselbach equation can be expressed as;

pH = pKa + log( [A⁻] / [HA])

the weak acid is dissociated into H⁺ and A⁻ ions in the solution.

now the conjugate base of the weak acid HA is

HA(aq) {weak acid}     H⁺(aq)  +  A⁻(aq) {conjugate base}

so now we calculate the value of Kₐ as well as pH value by substituting the values of the concentrations into the equation;

pKₐ = -logKₐ

pKₐ = -log ( 7.4×10⁻⁵ )

pKₐ = 4.13

now thw pH is

pH = pKₐ  + log( [A⁻] / [HA])

pH = 4.13 + log( [0.540] / [0.220])

pH = 4.13 + 0.3899

pH = 4.5199 = 4.52

Therefore the pH of the solution is 4.52

6 0
3 years ago
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