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SVETLANKA909090 [29]

3 years ago

6

Draw the alkyne formed when 1,1-dichloro-4-methylpentane is treated with an excess of a strong base such as sodium amide. (give

the neutralized alkyne product, not the acetylide salt.)

1 answer:

alisha [4.7K]3 years ago

4 0

Co2 is correct buddy

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A 0.200 g sample of unknown metal x is dropped into hydrochloride acid and realeases 80.3 mL of hydrogen gas at STP using ideal

s344n2d4d5 [400]

Answer:

The number of mole of the unknown metal is 3.58×10¯³ mole

Explanation:

We'll begin by calculating the number of mole hydrogen gas, H2 that will occupy 80.3 mL at stp.

This is illustrated below:

Recall:

1 mole of any occupy 22.4L or 22400 mL at stp.

1 mole of H2 occupies 22400 mL at stp.

Therefore, Xmol of H2 will occupy 80.3 mL at stp i.e

Xmol of H2 = 80.3/22400

Xmol of H2 = 3.58×10¯³ mole

Therefore, 3.58×10¯³ mole of Hydrogen gas was released.

Now, we can determine the mole of the unknown metal as follow:

The balanced equation for the reaction is given below:

X + 2HCl —> XCl2 + H2

From the balanced equation above,

1 mole of the unknown metal reacted to produce 1 mole of H2.

Therefore, 3.58×10¯³ mole of the unknown metal will also react to produce 3.58×10¯³ mole of H2.

Therefore, the number of mole of the unknown compound is 3.58×10¯³ mole.

5 0

3 years ago

Consider the electrolysis of aqueous agno3. Refer to the table of standard reduction potentials as needed. What should form at t

Valentin [98]

Answer:

Ag+

Explanation:

anode: 2AgNO3(l)⟶Ag(aq)+NO3(g)

8 0

3 years ago

The Sun always shines on half of the Moon. During a new moon, the Moon looks dark.

Sveta_85 [38]

You’re answer is C Explaination if the pit side was facing earth it wouldn’t be a new moon

8 0

3 years ago

Click the links to open the resources below. These resources will help you complete the assignment. Once you have created your f

White raven [17]

Answer:

I am not able to click anything above!

7 0

3 years ago

Wastewater discharged into a stream by a sugar refinery contains 3.40 g of sucrose (C12H22O11) per liter. A government-industry

Ipatiy [6.2K]

<u>Answer:</u> The pressure that must be applied to the apparatus is 0.239 atm

<u>Explanation:</u>

To calculate the osmotic pressure, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

or,

$\pi=i\times \frac{m_{solute}}{M_{solute}\times V_{solution}\text{ (in L)}}}\times RT$

where,

$\pi$ = osmotic pressure of the solution

i = Van't hoff factor = 1 (for non-electrolytes)

$m_{solute}$ = mass of sucrose = 3.40 g

$M_{solute}$ = molar mass of sucrose = 342.3 g/mol

$V_{solution}$ = Volume of solution = 1 L

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $20^oC=[20+273]K=293K$

Putting values in above equation, we get:

$\pi =1\times \frac{3.40g}{342.3g/mol\times 1}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 293K\\\\\pi =0.239atm$

Hence, the pressure that must be applied to the apparatus is 0.239 atm

3 0

3 years ago