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3 years ago

10

Name 3 Giant Covalent structures

2 answers:

KIM [24]3 years ago

8 0

SAND

GRAPHITE

DIAMOND

:)

8090 [49]3 years ago

6 0

1.Diamond

2.Graphite

3.Sand

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Our most recent mass extension happened about 65million years ago. Describe the effects of this event.

erik [133]

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To explain what caused this mass extinction, scientists have focused on events that would have altered our planet's climate in dramatic, powerful ways. The leading theory is that a huge asteroid or comet slammed into Earth 65 million years ago, blocking sunlight, changing the climate and setting off global wildfires.

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3 years ago

Which equation demonstrates that nuclear fusion forms elements that are heavier than helium?

WARRIOR [948]

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" B....Superscript 16 subscript 8 upper O plus superscript 4 subscript 2 upper H e right arrow superscript 20 subscript 10 upper N e."

if your on edgnuity

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3 years ago

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PLZ HELP!!!!!!<br> Which describes the molecule below????!?

Aleks [24]

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3 years ago

Calculate the mass of a body<br>Whose volume is<br>Is 2cm3 and<br> density is 520cm3

MrRissso [65]

Answer:

The answer is

<h2>1040g</h2>

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mass = density × volume

volume = 2cm³

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= 1040g

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3 years ago

A volume of 500.0 mL of 0.160 M NaOH is added to 585 mL of 0.200 M weak acid ( K a = 1.28 × 10 − 5 ) . What is the pH of the res

Hitman42 [59]

Answer : The pH of the resulting buffer is, 5.22

Explanation : Given,

$K_a=1.28\times 10^{-5}$

First we have to calculate the moles of $NaOH\text{ and }HA$

$\text{Moles of }NaOH=\text{Concentration of }NaOH\times \text{Volume of solution}}=0.160M\times 0.500L=0.08mol$

and,

$\text{Moles of }HA=\text{Concentration of }HA\times \text{Volume of solution}}=0.200M\times 0.585L=0.117mol$

The balanced chemical reaction is:

$HA+(aq)+OH^-(aq)\rightarrow H_2O(l)+A^-(aq)$

Moles of HA left = 0.117 mol - 0.08 mol = 0.037 mol

Moles of $A^-$ = 0.08 mol

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (1.28\times 10^{-5})$

$pK_a=5-\log (1.28)$

$pK_a=4.89$

Now we have to calculate the pH of buffer.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[A^-]}{[HA]}$

Now put all the given values in this expression, we get:

$pH=4.89+\log (\frac{0.08}{0.037})$

$pH=5.22$

Thus, the pH of the resulting buffer is, 5.22

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4 years ago