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Mamont248 [21]
3 years ago
6

How can you convert diamond into graphite ?​

Chemistry
2 answers:
Setler79 [48]3 years ago
4 0

Answer:

<em>➢</em><em>There are different ways to trigger the conversion of diamond to graphite, for instance by simply heating the diamond under exclusion of oxygen or even with an aimed mechanical stroke.</em>

Explanation:

<em>hope</em><em> it</em><em> will</em><em> help</em><em> you</em><em> have</em><em> a</em><em> great</em><em> day</em><em> bye</em><em> and</em><em> Mark</em><em> brainlist</em><em> if</em><em> the</em><em> answer</em><em> is</em><em> correct</em>

<em>kai6417</em>

<em>#</em><em>c</em><em>a</em><em>r</em><em>r</em><em>y</em><em> </em><em>on </em><em>learning</em>

Alexxx [7]3 years ago
3 0
By using ultra-short flashes of an x-ray laser
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In order to wash out the impurity, you looked up some sketchy method from 1952 which Hank Pym wrote during his Ph. D. under E. J
Eva8 [605]

Answer:

516.24

Explanation:

The solved solution is in the attached document.

8 0
3 years ago
What is the name for CH 3 CH 2 COCHCH 3 CH(CH 3 ) 2 ?
Paraphin [41]

Answer:

Butanoic acid

Explanation:

The IUPAC name of CH3CH2CH2COOH is:

The IUPAC name for a given compound is Butanoic acid.

4 0
3 years ago
How many molecules of Ca are found in a sample with 0.2 mols?
DochEvi [55]

Answer:

\boxed{1.2 \times 10^{23}\text{ atoms}}

Explanation:

6.023 × 10²³ atoms of Ca are in 1 mol of Ca

\text{No. of atoms} = \text{0.2 mol} \times \dfrac{6.023 \times 10^{23}\text{atoms }}{\text{1 mol }} = \mathbf{1.2 \times 10^{23}} \textbf{ atoms}}\\\\\text{There are }\boxed{\mathbf{1.2 \times 10^{23}} \textbf{ atoms}} \text{ atoms in 0.20 mol of Ca}

6 0
3 years ago
Which of the following compounds is always part of an aqueous solution?
jolli1 [7]
A.water.
An aqueous solution is a solution in water.
Hope this helped :)
5 0
3 years ago
Read 2 more answers
Sodium hydrogen carbonate NaHCO3, also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. Ac
olga2289 [7]

Answer:

1.4952 grams of sodium bicarbonate she  would need to ingest to neutralize this much HCl.

Explanation:

Moles (n)=Molarity(M)\times Volume (L)

Moles of hydrochloric acid = n

Volume of hydrochloric acid solution = 200.0 mL = 0.200 L

Molarity of the hydrochloric acid = 0.089 M

n=0.089 M\times 0.200 L=0.0178 mol of HCL

HCl(aq)+NaHCO_3(aq)\rightarrow NaCl(aq)+H_2O(l)+CO_2(g)

According to reaction, 1 mole of HCl is neutralized by 1 mole of sodium bicarbonate.

Then 0.0178 moles of HCl wil be neutralized by :

\frac{1}{1}\times 0.0178 mol=0.0178 mol of sodium bicarbonate

Mass of 0.0178 moles of sodium bicarbonate:

0.0178 mol × 72 g/mol = 1.4952 g

1.4952 grams of sodium bicarbonate she  would need to ingest to neutralize this much HCl.

8 0
3 years ago
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