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VashaNatasha [74]
3 years ago
9

The structure of butanoic acid

Chemistry
1 answer:
makvit [3.9K]3 years ago
7 0

answer on here fjchcjfjdnc.com

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The structural formula of 2-methyl-propane<br>-2-ol​
daser333 [38]

Answer:

ch3c(ch3)(oh)ch3

Explanation:

that should be the answer

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3 years ago
Organic chemistry is the study of chemicals that no contain carbon true or false
Artist 52 [7]
False
because carbon is very important in organic chemistry, u need to study that
4 0
4 years ago
A 18.00 g g milk chocolate bar is found to contain 11.00 g g of sugar.1. How many milligrams of sugar does the milk chocolate ba
horrorfan [7]

Answer:

1. 1.100x10⁴mg

2. 2444mg

Explanation:

1. The 18.00g milk chocolate bar contain 11.00g of sugar. In miligrams:

11.00g * (1000mg / 1g) = 1.100x10⁴mg

2. If a bar of 18.00g contain 11.00g of sugar, a bar of 4.000g will contain:}

4.000g bar * (11.00g / 18.00g)  = 2.444g of sugar.

In miligrams:

2.444g * (1000mg / 1g) = 2444mg

6 0
3 years ago
Please answer fast do in 5 mins Body System used for support, movement, structure and makes red blood cells
pochemuha

Explanation:

The Skeletal System (A).

7 0
3 years ago
A 0.2500 g sample of a compound known to contain carbon, hydrogen and oxygen undergoes complete combustion to produce 0.3664 gra
irga5000 [103]

Answer:

C₂H₅O₂

Explanation:

From the question given above, the following data were obtained:

Mass of compound = 0.25 g

Mass of CO₂ = 0.3664 g

Mass of H₂O = 0.15 g

Empirical formula =?

Next, we shall determine the mass of carbon, hydrogen and oxygen present in the compound. This can be obtained as follow:

For Carbon (C):

Mass of CO₂ = 0.3664 g

Molar mass of CO₂ = 12 + (2×16) = 44 g/mol

Mass of C = 12/44 × 0.3664

Mass of C = 0.1

For Hydrogen (H):

Mass of H₂O = 0.15 g

Molar mass of H₂O = (2×1) + 16 = 18 g/mol

Mass of H = 2/18 × 0.15

Mass of H = 0.02 g

For Oxygen (O):

Mass of C = 0.1 g

Mass of H = 0.02 g

Mass of compound = 0.25 g

Mass of O =?

Mass of O = (Mass of compound) – (Mass of C + Mass of H)

Mass of O = 0.25 – (0.1 + 0.02)

Mass of O = 0.25 –0.12

Mass of O = 0.13 g

Finally, we shall determine the empirical formula for the compound. This can be obtained as follow:

C = 0.1

H = 0.02

O = 0.13

Divide by their molar mass

C = 0.1 / 12 = 0.0083

H = 0.02 / 1 = 0.02

O = 0.13 / 16 = 0.0081

Divide by the smallest

C = 0.0083 / 0.0081 = 1

H = 0.02 / 0.0081 = 2.47

O = 0.008 / 0.008 = 1

Multiply by 2 to express in whole number

C = 1 × 2 = 2

H = 2.47 × 2 = 5

O = 1 × 2 = 2

Thus, the empirical formula for the compound is C₂H₅O₂

5 0
3 years ago
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