Identify balance chemical equation from the following

Almost. treating an acyl chloride with a grignard reagent tends to form (via two steps) an alcohol. what reagent forms the keton

PIT_PIT [208]

Reaction of Acyl Halides with Grignard reagent results in the formation of Ketones in first step. While in second step reaction of Grignard reagent with Ketones results in the formation of Tertiary Alcohols.

If you want to stop the reaction at Ketone stage then you are required to use another mild reactive organometallic compound. In our case we will use Organocuprates. Organocuprates are also known as Gilman Reagents. These reagents does not add to ketones, aldehydes and esters but they can add to acid halides to produce Ketones.

8 0

3 years ago

What are the empirical formula and empirical formula mass for C10H30O10?

AysviL [449]

Answer:

Empirical formula: CH₃O

Empirical formula mass = 31 g/mol

Explanation:

Data Given:

Molecular Formula = C₁₀H₃₀O₁₀

Empirical Formula = ?

Empirical Formula mass =

Solution

Empirical Formula:

Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.

So,

The ratio of the molecular formula should be divided by whole number to get the simplest ratio of molecule

As

C₁₀H₃₀O₁₀ Consist of 10 Carbon (C) atoms, 30 Hydrogen (H) atoms, and 10 Oxygen (O) atoms.

Now

Look at the ratio of these three atoms in the compound

C : H : O

10 : 30 : 10

Divide the ratio by two to get simplest ratio

C : H : O

10/10 : 30/10 : 10/10

1 : 3 : 1

So for the empirical formula the simplest ratio of carbon to hydrogen to oxygen is 1:3:1

So the empirical formula will be

Empirical formula of C₁₀H₃₀O₁₀ = CH₃O

Now

To find the empirical formula mass in g/mol

Formula mass:

Formula mass is the total sum of the atomic masses of all the atoms present in a formula unit.

**Note:

if we represent the molar mass of the empirical formula for one mol in grams then it is written as g/mol

So,

As the empirical formula of C₁₀H₃₀O₁₀ is CH₃O

Then Its empirical formula mass will be

CH₃O

Atomic Mass of C = 12

Atomic Mass of H = 3

Atomic Mass of O = 16

Total Molar mass of CH₃O

CH₃O = 12 + 3(1) + 16

CH₃O = 12 + 3 + 16

CH₃O = 31 g/mol

4 0

3 years ago

How do you find the molar mass for ba(NO3)2

Korvikt [17]

To find the molar mass of Ba(NO3)2, determine the molar masses of all the atoms that form it. The Molar mass for Barium nitrate is 261.337 g/mol.

8 0

3 years ago

If 1.02 g of nickel reacted with 750. mL of 0.112 M hydrobromic acid, how much of each will be present at the end of the reactio

kati45 [8]

Answer:

35.1% is percent yield

Explanation:

Full question: Assume no volume change. If you formed 0.0910 atm of gas, what is the percent yield?

The reaction that is occurring is:

Ni + 3HBr → NiBr₃ + 3/2H₂(g)

First, we will determine moles of Ni and HBr to determine limiting reactant and theoretical yield

Using ideal gas law, we can determine the moles of hydrogen formed. Thus, we can find percent yield:

Moles Ni (Molar mass: 58.69g/mol):

1.02g * (1mol / 58.69g) = 0.01738moles Ni

Moles HBr:

0.750L * (0.112mol/L) = 0.084 moles of HBr.

For a complete reaction of the 0.084 moles of HBr you need:

0.084mol HBr * (1 mole Ni / 3 moles HBr) = 0.028 moles of Ni.

As there are just 0.01738 moles of Ni, the Ni is limiting reactant. Assuming a theoretical yield, moles of H₂ produced are:

0.01738moles Ni * (3/2 H₂ / 1 mol Ni) = 0.02607 moles H₂

Now, moles of H₂ produced are:

PV = nRT

PV/RT = n

Where P is pressure (0.0910atm)

V is volume (2.50L)

R is gas constant (0.082atmL/molK)

T is absolute temperature in Kelvin (30°C + 273.15 = 303.15K)

And n are moles

PV/RT = n

0.0910atm*2.50L/0.082atmL/molK*303.15K = n

0.00915 moles = n

And percent yield (Produced moles / Theoretical moles * 100) is:

0.00915 moles / 0.02607moles =

<h3>35.1% is percent yield</h3>

8 0

3 years ago

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m

koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

$R_{in}$ $= \frac{0.04}{100}*2000$