Ask question

Ask question

All categories

grandymaker [24]

3 years ago

5

Which element easily loses one electron to form a positive ion? A. Magnesium (Mg) B. Lithium (Li) C. Nitrogen (N) D. Fluorine (F

)

1 answer:

gladu [14]3 years ago

3 0

Answer: Lithium (Li)

Explanation:

Once it loses its only valence electron, there’s no longer a negative charge.

You might be interested in

Which of the following is an important intermediate in the mechanism of the reaction<br>in the box?

Dennis_Churaev [7]

Answer:

Explanation:

.

3 0

3 years ago

A solution was prepared by mixing 20.00 mL of 0.100 M and 120.00 mL of 0.200 M. Calculate the molarity of the final solution of

Marat540 [252]

Answer:

0.186M

Explanation:

First, we need to obtain the moles of nitric acid that are given for each solution. Then, we need to divide these moles in total volume (120mL + 20mL = 140mL = 0.140L) to obtain molarity:

<em>Moles Nitric acid:</em>

0.0200L * (0.100mol / L) = 0.00200 moles

0.120L * (0.200mol / L)= 0.02400 moles

Total moles: 0.02400moles + 0.00200moles = 0.026 moles of nitric acid

Molarity: 0.026 moles / 0.140L

<h3>0.186M</h3>

6 0

3 years ago

2-butanone is converted into 3-methyl-3-hexanol using a grignard reagent prepared from 1-bromopropane and magnesium metal in thf

Andre45 [30]

Answer:

Here's what I get.

Explanation:

At the end of the reaction you will have a solution of the alcohol in THF.

The microdistillation procedure will vary, depending on the specific apparatus you are using, but here is a typical procedure.

Transfer the solution to a conical vial.
Add a boiling stone.
Attach a Hickman head (shown below) and condenser.
Place the assembly in in the appropriate hole of an aluminium block on top of a hotplate stirrer.
Begin stirring and heating at a low level so the THF (bp 63 °C) can distill slowly.
Use a Pasteur pipet to withdraw the THF as needed.
When all the THF has been removed, raise the temperature of the Al block and distill the alcohol (bp 143 °C).

7 0

3 years ago

Suppose you have just added 200.0 ml of a solution containing 0.5000 moles of acetic acid per liter to 100.0 ml of 0.5000 M NaOH

uranmaximum [27]

Answer:

The final pH is 3.80

Explanation:

Step 1: Data given

Volume of acetic acid = 200.0 mL = 0.200 L

Number of moles acetic acid = 0.5000 moles

Volume of NaOH = 100.0 mL = 0.100 L

Molarity of NaOH = 0.500 M

Ka of acetic acid = 1.770 * 10^-5

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles

moles = molarity * volume

Moles NaOH = 0.500 M * 0.100 L

Moles NaOH = 0.0500 moles

Step 4: Calculate the limiting reactant

For 1 mol CH3COOH we need 1 mol NaOH to produce 1 mol CH3COONa and 2 moles H2O

NaOH is the limiting reactant. It will completely be consumed (0.0500 moles). CH3COOH is in excess. There will react 0.0500 moles . There will remain 0.500 - 0.0500 = 0.450 moles

There will be produced 0.0500 moles CH3COONa

Step 5: Calculate the total volume

Total volume = 200.0 mL + 100.0 mL = 300.0 mL

Total volume = 0.300 L

Step 6: Calculate molarity

Molarity = moles / volume

[CH3COOH] = 0.450 moles / 0.300 L

[CH3COOH] = 1.5 M

[CH3COONa] = 0.0500 moles / 0.300 L

[CH3COONa]= 0.167 M

Step 7: Calculate pH

pH = pKa + log[A-]/ [HA]

pH = -log(1.77*10^-5) + log (0.167/ 1.5)

pH = 4.75 + log (0.167/1.5)

pH = 3.80

The final pH is 3.80

7 0

3 years ago

5/3 * 2/3 what's the answer

Contact [7]

5 x 2 = 10

3 x 3 = 9

10/9, or 1 1/9 is your answer

hope this helps

6 0

3 years ago