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3 years ago

Why are the densities of corn syrup and gasoline expressed as a range of values?

Chemistry

2 answers:

SVETLANKA909090 [29]3 years ago

6 0

Answer:

The answer is because they are not a substance, but a mixture of organic compounds.

Explanation:

A range of values is used because they are not a substance, but a mixture of organic compounds that has a range of molar masses. Corn syrup contains glucose in not less than 70% by weight of total solids and a reducing sugar content of 20% by weight on a dry basis. The composition of gasoline depends on the source from which it is extracted, therefore it has different molar masses. A gasoline with a long carbon chain is heavy and has a higher mass for a fixed volume, so its density will increase. Any change affecting the product will lead to an increase or decrease in density.

Have a nice day!

Maslowich3 years ago

4 0

A range of values is given for the density of gasoline and corn syrup could be because these substances changes their volumes along with the changes in temperature which will result to a change in density as well. Hope this answers the question.

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if 1.386 g of mg ribbon combusts to form 2.309 g of oxide product, calculate the experimental mass percent of oxygen from this d

Pavlova-9 [17]

1.386 g of Mg ribbon combusts to form 2.309 g of oxide product. The mass percent of oxygen in the oxide is 40.0 %.

Let's consider the reaction for the combustion of Mg.

Mg + 1/2 O₂ ⇒ MgO

1.386 g of Mg combusts to form 2.309 g of MgO. We want to determine the mass of oxygen in MgO. According to Lavoisier's law of conservation of mass, matter is not created nor destroyed over the course of a chemical reaction. Then, the mass of Mg in the reactants is equal to the mass of Mg in MgO. The mass of the magnesium oxide is the sum of the masses of magnesium and oxygen. The <u>mass of oxygen in the oxide</u> is:

$mMgO = mMg + mO\\mO = mMgO - mMg = 2.309 g - 1.386 g = 0.923 g$

We can calculate the mass percent of O in MgO using the following expression.

$\% O = \frac{mO}{mMgO} \times 100\% = \frac{0.923 g}{2.309g} \times 100\% = 40.0 \%$

You can learn more about mass percent here: brainly.com/question/14990953

3 0

1 year ago

What type of compound is calcium oxide

DIA [1.3K]

A carbonate because the oxygen neutralizes the reactive coding of the calcium in its original form.

5 0

3 years ago

Which of the following is the correct name for the compound mgbr2? magnesium bromate magnesium bromide magnesium dibromide magne

Sergeeva-Olga [200]

Correct answer is magnesium bromide. This is an ionic compound with metal forming a positive ion - K+ and halogen forming a negative ion - Br-. When group 7 element form ions they have a suffix -ide. Bromine is the element and when it forms a negative ion bromine is called bromide with the suffix.
Magnesium gives 2 electrons and bromine can take only 1 electron, therefore 2 bromine atoms are needed. Therefore magnesium forms ionic bonds with 2 bromine atoms.
The compound is called magnesium bromide

6 0

3 years ago

50 points!! Brainliest if correct!!

Luda [366]

The answer is C. The answer is C because if u increase the surface area, the more reactants u will get. and if u get more The reactants will move faster. Hoped that Helped!:-)

4 0

3 years ago

37.9 grams of an unknown substance undergoes a temperature increase of

jenyasd209 [6]

Answer:

1.023 J / g °C

Explanation:

m = 37.9 grams

ΔT = 25.0*C

H = 969 J

c = ?

The equation relating these equation is;

H = mcΔT

making c subject of formulae;

c = H / mΔT

c = 969 J / (37.9 g * 25.0*C)

Upon solving;

c = 1.023 J / g °C

4 0

3 years ago