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3 years ago

What is a substance combination of different atoms?

2 answers:

5 0

Every combination of atoms is a molecule. A compound is a molecule made of atoms from different elements. All compounds are molecules, but not all molecules are compounds

anastassius [24]3 years ago

3 0

Answer:

Brainliest-

Explanation:

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Classify each of the following as a homogeneous or heterogeneous substance.

Mnenie [13.5K]

Answer:

a). iron ore - is a heterogeneous substance.

b). quartz - is a homogeneous substance.

c). granite - is a heterogeneous substance.

d). energy drink - is a heterogeneous substance.

e). oil-and-vinegar salad dressing - is a heterogeneous substance.

f). salt - is a homogeneous substance.

g). rainwater - is a heterogeneous substance.

h). nitrogen - is a homogeneous substance.

Explanation:

a). iron ore - is a heterogeneous substance as iron core contains iron and other impurities

b). quartz - is a homogeneous substance as it is only made up of carbon

c). granite - is a heterogeneous substance as it is made up of multiple minerals.

d). energy drink - is a heterogeneous substance as it contains various substances mixed in it

e). oil-and-vinegar salad dressing - is a heterogeneous substance as oil and vinegar itself are different compounds

f). salt - is a homogeneous substance as it only contains one type of molecule

g). rainwater - is a heterogeneous substance as it contains water and other minerals

h). nitrogen - is a homogeneous substance as it is only made of one type of molecule.

5 0

1 year ago

A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

Answer: The mass of glucose in final solution is 0.420 grams

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$

Hence, the mass of glucose in final solution is 0.420 grams

3 0

3 years ago

An atom of aluminum in the ground state and an atom of gallium in the ground state have the same

Kobotan [32]

(4) total number of valence electrons, because they exist in the same group.

3 0

4 years ago

Predict the effect of adding a non competitive inhibitor to the reaction mixture on the rate of reaction at a high substrate con

Darya [45]

Answer:

A noncompetitive inhibitor can only bind to an enzyme with or without a substrate at several places at a particular point in time

Explanation:

this is because It changes the conformation of an enzyme as well as its active site, which makes the substrate unable to bind to the enzyme effectively so that the efficiency of the enzyme decreases. A noncompetitive inhibitor binds to the enzyme away from the active site, altering/distorting the shape of the enzyme so that even if the substrate can bind, the active site functions less effectively and most of the time also the inhibitor is reversible

8 0

4 years ago

Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$ ............(1)