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3 years ago

Could a mixture be made up of only elements and no compounds? Explain.

2 answers:

5 0

No. that's why it is called a mixture. If it were called a compound it could be made up of different elements, but a mixture is irreversible and I made up of elements and compounds.

Art [367]3 years ago

3 0

Yes. A mixture in definition is a combination of two or more elements or compounds. Hence in combining these elements, mixture is another set of chemical bonding, separately and divergent. Compounds are not necessary to form mixtures. There is an “or” in the said definition and certain authors to support this claim. These includes examples that can state clearly this argument, table salt as a mixture of independent elements, Na and Cl. Brass is composed of zinc and copper. Steel is from carbon and iron. Lastly, you can have gold, nickel and palladium or platinum to form white gold.

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What are the four characteristics of plants

UNO [17]

Start of the food chain, one of the keys to survival and, one of the most natural beauties

8 0

4 years ago

Solid chromium (III) reacts with oxygen gas to form solid Cr2O3. What is this type of reaction?

mafiozo [28]

Answer:

.081 g of O2

Explanation:

4Cr + 3O2 -----> 2Cr2O3

.175 g Cr x [1 mole / 52.0 g] x [2 moles Cr2O3 / 4 moles Cr] x [152 g / 1 mole] = .256 g of Cr2O3

.175 g Cr x [1 mole / 52.0 g] x [3 moles O2 / 4 moles Cr] x [32 g / 1 mole] = .081 g of O2

5 0

3 years ago

A certain first-order reaction (A→productsA→products) has a rate constant of 9.90×10−3 s−1s−1 at 45 ∘C∘C. How many minutes does

coldgirl [10]

Answer:

4.66667 minutes

Explanation:

Rate constant, k = 9.90×10−3 s−1

Time = ?

Initial concentration, [A]o = 100

Final concentration, [A] = 6.25

The integral rate law for first order reactions is given as;

ln[A] = ln[A]o − kt

kt = ln[A]o - ln[A]

t = ( ln[A]o - ln[A]) / k

t = [ln(100) - ln(6.25)] / 9.90×10−3

t = 2.77 / 9.90×10−3

t = 0.28006 ×103

t = 280 seconds

t = 4.66667 minutes (Upon conversion by dividing by 60)

4 0

3 years ago

What is the independent variable of the sentence "if diet affects weight, then people who eat less will lose weight."?

Citrus2011 [14]

The quantity or amount of food that people consume would be categorized as the independent variable.

7 0

3 years ago

N2+3H2---> 2NH3

irakobra [83]

Hello!

A) How many grams of hydrogen are necessary to react completely with 50.0g of nitrogen in the above reaction?

The balanced chemical reaction is the following one:

N₂(g) + 3H₂(g) → 2NH₃(g)

To calculate the amount of hydrogen necessary we will use the following conversion factor to go from grams of nitrogen to grams of hydrogen:

$50 gN_2* \frac{1 mol N_2}{28,013 g N_2}* \frac{3 moles H_2}{1 mol N_2}* \frac{2,016 g H_2}{1 mol H_2}=10,7950 g H_2$

So, 10,7950 grams of H₂ are required to react with 50 g of nitrogen

B) How many grams of ammonia are produced in the reaction from the previous problem?

The balanced chemical reaction again is the following one:

N₂ + 3H₂ → 2NH₃

To calculate the amount of ammonia produced we will use the following conversion factor to go from grams of nitrogen to grams of ammonia:

$50 gN_2* \frac{1 mol N_2}{28,013 g N_2}* \frac{2 moles NH_3}{1 mol N_2}* \frac{17,031 g NH_3}{1 mol NH_3}=60,7968 g NH_3$

So, 60,7968 grams of NH₃ are produced from 50 g of nitrogen

C) How many grams of silver chloride are produced from 5.0g of silver nitrate reacting with an excess of barium chloride?

The balanced chemical equation for the reaction is the following one:

2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂

To calculate the mass of Silver Chloride produced we will use the following conversion factor to go from grams of Silver Nitrate to grams of Silver Chloride:

$5 g AgNO_3* \frac{1 mol AgNO_3}{169,87 g AgNO_3}* \frac{2 moles AgCl}{2 moles AgNO_3}* \frac{143,32 g AgCl}{1 mol AgCl}=4,2185 g AgCl$

So, 4,219 g of AgCl are produced from the reaction of 5 g of AgNO₃

D) How much barium chloride is necessary to react with the silver nitrate in the previous problem?

The balanced chemical equation for the reaction is the following one:

2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂

To calculate the mass of Barium Chloride necessary we will use the following conversion factor to go from grams of Silver Nitrate to grams of Barium Chloride:

$5 g AgNO_3* \frac{1 mol AgNO_3}{169,87 g AgNO_3}* \frac{1 mol BaCl_2}{2 moles AgNO_3}* \frac{208,23 g BaCl_2}{1 mol BaCl_2}=3,08 g BaCl_2$

So, 3,08 g of BaCl₂ are produced from the reaction of 5 g of AgNO₃

Have a nice day!

4 0

3 years ago