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wel
3 years ago
14

Hi......................... HELLO

Chemistry
2 answers:
GREYUIT [131]3 years ago
7 0

Answer:

Hii hello how are you....

PtichkaEL [24]3 years ago
6 0

Answer:

hi

Explanation:

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2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)<br> how many atoms are in 3.2 moles of KCl?
Gekata [30.6K]

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If u use math why It will give u the answer like 3x3=9

Explanation:

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2 years ago
If electricity comes from electrons, does morality come from morons?
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No morality come from each individual person.
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3 years ago
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CaC2 + 2H2O ⟶ C2H2 + Ca(OH)2
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0.499 mol

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Amount of Calcium hydroxide Is same

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3 years ago
What does an energy diagram illustrate? Identify what the "hill" on the energy curve and the positions of the reactants and prod
k0ka [10]
An energy diagram of a chemical reactions illustrates the changes of energy as the chemical reactions advances.

At first, the energy in the diagram is the energy of the reactants.

As the reaction goes forward, the reactants start to react forming a transition compound, with a maximum energy level on the graph. This is the hill. So the hill represents the Activation Energy.

After that, the energy starts to decrease and at the end you have the energy of the products, which may be higher or lower than the initial energy of the reactants, depending upon whether the reaction is exothermic or endothermic.

For exothermic reactions the energy level of the products is lower than the energy level of the reactants, while for endothermic reactions the energy level of the products is higher than the energy level of the reactants.
5 0
3 years ago
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The solubility of acetanilide is 12.8 g in 100 mL of ethanol at 0 ∘C, and 46.4 g in 100 mL of ethanol at 60 ∘C. What is the maxi
weqwewe [10]

Answer: 72.41% and 26.90% respectively.

Explanation:

At 60°C, you can dissolve 46.4g of acetanilide in 100mL of ethanol. If you lower the temperature, at 0°C, you can dissolve just 12.8g, which means (46.4g-12.8g)=33.6g of acetanilide must have precipitated from the solution.

We can calculate recovery as:

\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{33.6\ g}{46.4\ g}*100=72.41\%

So the answer to the first question is 72.41%.

For the second part just use the same formula, the mass of the precipitate is the final mass minus the initial mass, (171mg-125mg)=46mg.

\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{46\ mg}{171\ mg}*100=26.90\%

So the answer to the second question is 26.90%.

3 0
3 years ago
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