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krok68 [10]
4 years ago
14

Lewis structure for Potassium oxide And Sodium oxide, please.

Chemistry
1 answer:
Maslowich4 years ago
4 0
K-O-K (put four dots around the O)
Na-O-Na (put four dots around the O)
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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.

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3 years ago
Can someone solve this plz<br> βeta decay<br> carbon atomic mass 14 <br> atomic # 6
nordsb [41]

Answer:

14/6C your welcome

Explanation:

8 0
3 years ago
Nickel (iii) cyanide + aluminum permanganate product?
Wewaii [24]

Answer:

Nickel(II) cyanide is an inorganic compound with a chemical formula Ni(CN)₂

Explanation:

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5 0
3 years ago
For the following reaction, 13.1 grams of glucose (C6H12O6) are allowed to react with 10.6 grams of oxygen gas. glucose (C6H12O6
mart [117]

Answer:

There will be formed 14.58 grams of CO2

O2 is the limiting reagent

There will remain 3.151 grams of glucose

Explanation:

Step 1: Data given

Mass of glucose = 13.1 grams

molar mass of glucose 180.156 g/mol

Mass of oxygen = 10.6grams

molar mass of oxygen = 32 g/mol

<u>Step 2</u>: The balanced equation

C6H12O6 + 6 O2 → 6H2O + 6CO2

<u>Step 3</u>: Calculate moles of glucose

Moles of glucose = mass glucose / molar mass glucose

Moles of glucose = 13.1 grams / 180.156 g/mol

Moles of glucose = 0.0727 moles

Step 4: Calculate moles of oxygen

Moles O2 = 10.6 grams / 32 g/mol

Moles O2 = 0.33125 moles

Step 5: Calculate the limiting reactant

For 1 mol of glucose , we need 6 moles of O2 to produce 6 moles of H2O and 6 moles of CO2

Oxygen is the limiting reactant. It will completely be consumed ( 0.33125 moles).

Glucose is in excess. There will be consumed 0.33125/6 = 0.05521 moles

There will remain 0.0727 - 0.05521 = 0.01749 moles

This is 0.01749 moles * 180.156 g/mol = 3.151 grams

Step 6: Calculate moles of CO2

For 1 mol of glucose , we need 6 moles of O2 to produce 6 moles of H2O and 6 moles of CO2

For 0.33125 moles of O2 we'll get 0.33125 moles of CO2 produced

Step 7: Calculate mass of CO2

Mass of CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.33125 moles * 44.01 g/mol

Mass CO2 = 14.58 grams

7 0
3 years ago
What conclusions can be drawn regarding the molecule C6H12O6?
tekilochka [14]
4. The molecule is organic; it is a carbohydrate and 2. Sodium and aluminum have similar chemical properties because they are in the same period.
4 0
3 years ago
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