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3 years ago

How many oxygen atoms are present in 3 grams of glucose

Chemistry

1 answer:

Kaylis [27]3 years ago

5 0

Answer:

Because CLEARLY, each mole of glucose, C6H12O6 contains 6⋅mol oxygen atoms.

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draw the structure of two acyclic compounds with 3 or more carbons which exhibits one singlet in the 1H-NMR spectrum

loris [4]

Answer:

attached below

Explanation:

Structure of two acyclic compounds with 3 or more carbons that exhibits one singlet in 1H-NMR spectrum

a) Acetone CH₃COCH₃

Attached below is the structure

b) But-2-yne (CH₃C)₂

Attached below is the structure

5 0

3 years ago

ch sulfide (ZnS) occurs in the zinc blende crystal structure, (a) If 254 g of ZnS contains 170 g of Zn, what is the mass ratio o

Aleonysh [2.5K]

Answer:

The mass ratio of zinc to sulfide is 85:42.

2.5559 kg of Zn are in 3.82 kg of ZnS.

Explanation:

a) Mass of zinc sulfide = 254 g

Mass of zinc in a zinc sulfide sample = 170 g

Mass of sulfide in zinc sulfide sample = x

254 g = 170 g+ x

x = 84 g

The mass ratio of zinc to sulfide:

$\frac{170 g}{84 g}=\frac{85}{42}$

b) Mass of zincsulfide sample = 3.83 kg

The mass ratio of zinc to sulfide is 85:42.

Let the mass of zinc and sulfide be 85x and 42x respectively:

85 x+ 42 x=3.82 kg

x =0.03007 kg

Mass of an zinc= 85x=85 × 0.03007 kg= 2.5559 kg

8 0

3 years ago

To find the number of atoms look at the number known as the________. Please help.

Rufina [12.5K]

Answer:

molar mass or AMU

Explanation:

4 0

2 years ago

Using the reagents below, list in order (by letter, no period) those necessary to transform 1-chlorobutane into 1-butyne. Note:

Andru [333]

Answer:

gde

Explanation:

We are attempting to synthesize 1-butyne from 1-chlorobutane. Since 1-chlorobutane is a primary alkyl halide, 1-butene is formed when 1-chlorobutane is reacted with a bulky base such as t -BuOK or t -BuOH in presence of strong heat. This is an E2 reaction.

Secondly, the 1-butene is reacted with bromine in carbon tetrachloride. The vicinal dihalide (1,2-dibromobutane) is formed. This can now undergo further elimination reactions in the presence of sodamide and strong heat to yield 1-butyne which is the desired product. These reactions involve the elimination of the first HBr molecule to give an alkenyl bromide. A second elimination now gives the terminal alkyne.

3 0

3 years ago

Could someone help me with this?

Varvara68 [4.7K]

Solving part-1 only

KMnO_4

Transition metal is Manganese (Mn)

Actually it's the oxidation number of Mn

Let's find how?

$\\ \tt\Rrightarrow x+1+4(-2)=0$

$\\ \tt\Rrightarrow x+1-8=0$

$\\ \tt\Rrightarrow x-7=0$

$\\ \tt\Rrightarrow x=+7$

x is the oxidation number

Purple as per the color of potassium permanganate

$\boxed{\begin{array}{c|c|c}\boxed{\bf Tube} &\boxed{\bf Charge} &\boxed{\bf No\:of\; electrons\: loss}\\ \sf 2 &\sf +6 &\sf 6e^-\\ \sf 3& \sf +2 &\sf 2e- \\ \sf 4 &\sf 4 &\sf 4e^-\end{array}}$

7 0

2 years ago