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Ber [7]
3 years ago
15

What type of ions have names ending in ide?

Chemistry
1 answer:
Anon25 [30]3 years ago
7 0
The answer would be C.
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Predict the product formed when the compound shown below undergoes a reaction with mcpba in ch2cl2. mcpba is meta-chloroperoxybe
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Alkenes on reaction with meta-chloroperoxybenzoic acid (MCPBA ) produces epoxides. When styrene is reacted with mCPBA it gives 2-phenyloxirane as shown below,

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How many moles of oxygen are present in 16 g of oxygen gas​
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Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as
zubka84 [21]

Answer:

Explanation:

MnO₂(s) + 4 HCl(aq)  = MnCl₂(aq) + 2 H₂O(l) + Cl₂

87 g                                                                     22.4 x 10³ mL

volume of given chlorine gas at NTP or at 760 Torr and 273 K

=  175 x ( 273 + 25 ) x 715 / (273 x 760 )

= 179.71 mL

22.4 x 10³ mL of chlorine requires 87 g of MnO₂

179.4 mL of chlorine will require    87 x 179.4 / 22.4 x 10³ g

= 696.77 x 10⁻³ g

= 696.77 mg .

6 0
3 years ago
What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m
djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the  solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of  sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL  (1cm³  = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³  mol / 0.4327 L = 0.0075  M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x  432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for  volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

8 0
3 years ago
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