What are the products of C6 H12 + 9O2 ?

nirvana33 [79]

Answer:

I think the answer is D. Temperature affects only the rate of reaction.

7 0

3 years ago

Your friend the archaeologist, who studies the ancient human past, has asked you to find a numerical date for some hominid artif

natulia [17]

Answer:

I think it the carbon 14 which is a half life of 5730 because in 4.5billion years ago I don't think any human being existed on earth I'm not sure I hope that I helped you and good luck .

3 0

2 years ago

Where are atoms located? (check all that apply)*

Jobisdone [24]

I think it’s 2 hope that helped

4 0

2 years ago

In substitution reactions, (CH3)3C-I reacts at the same rate with Br- and Cl- even though Br- is a more reactive nucleophile tha

Kamila [148]

Answer:

A. (CH3)3C-I reacts by SN1 mechanism whose rate is independent of nucleophile reactivity.

Explanation:

We must recall that (CH3)3C-I is a tertiary alkyl halide. Tertiary alkyl halides preferentially undergo substitution reaction via SN1 mechanism.

In SN1 mechanism, the rate of reaction depends solely on the concentration of the alkyl halide (unimolecular mechanism) and is independent of the concentration of the nucleophile. As a result of this, both Br^- and Cl^- react at the same rate.

5 0

3 years ago

If you have 16 g of manganese (II) nitrate tetrahydrate, how much water is required to prepare 0.16 M solution from this amount

Schach [20]

Answer: The amount of water required to prepare given amount of salt is 398.4 mL

Explanation:

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 0.16 M

Given mass of manganese (II) nitrate tetrahydrate = 16 g

Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol

Putting values in above equation, we get:

$0.16M=\frac{16\times 1000}{251\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{16\times 1000}{251\times 0.16}=398.4mL$

Volume of water = Volume of solution = 398.4 mL

Hence, the amount of water required to prepare given amount of salt is 398.4 mL

4 0

3 years ago