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Chemistry

1 answer:

Semenov [28]3 years ago

6 0

Answer:

1. 71 and 29

2. 97 and 3

3. 2.1 and 1.2

Explanation:

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Define saturated and unsaturated fats

Sphinxa [80]

Answer:

<h3><u>Saturated and unsaturated</u>:-</h3>

Are a form of fat in which all or most of the fatty acid chains are single bonds. Glycerol and fatty acids are the two types of smaller molecules that make up fat.

<h3><u>Saturated fat is found in:</u></h3>

cakes

sausages
cheese
butter

<h3><u>Examples of unsaturated fats:- </u></h3>

Olive
Nuts(almonds, hazelnuts)
Seeds(pumpkin and sesame seeds)

hope it helps...

6 0

3 years ago

Carbon disulfide is prepared industrially by reacting carbon with sulfur dioxide according to the above equation. If 70.8 g of c

Rus_ich [418]

Answer:

1.18 moles of CS₂ are produced by the reaction.

Explanation:

We present the reaction:

5C + 2SO₂ → CS₂ + 4CO

5 moles of carbon react to 2 moles of sulfur dioxide in order to produce 1 mol of carbon disulfide and 4 moles of carbon monoxide.

As we do not have data from the SO₂, we assume this as the excess reagent. We convert the mass of carbon to moles:

70.8 g / 12 g/mol = 5.9 moles

Ratio is 5:1, so 5 moles of carbon react to produce 1 mol of CS₂

Then, 5.9 moles will produce (5.9 . 1) / 5 = 1.18 moles

8 0

3 years ago

3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$