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bekas [8.4K]
3 years ago
15

What does avogadro’s number represent?

Chemistry
1 answer:
Harman [31]3 years ago
5 0
Number of elementary particles
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I need help fast please <br> What is ATP?
Jobisdone [24]

Answer:

Adenosine triphosphate is a complex organic chemical that provides energy to drive many processes in living cells, e.g. muscle contraction, nerve impulse propagation, and chemical synthesis. Found in all forms of life, ATP is often referred to as the "molecular unit of currency" of intracellular energy transfer. Wikipedia

Formula: C10H16N5O13P3

Molar mass: 507.18 g/mol

IUPAC ID: [[(2R,3S,4R,5R)-5-(6-aminopurin-9-yl)-3,4-dihydroxyoxolan-2-yl]methoxy-hydroxyphosphoryl] phosphono hydrogen phosphate

Acidity (pKa): 6.5

Soluble in: Water

UV-vis (λmax): 259 nm

Explanation:

8 0
4 years ago
Read 2 more answers
The digestion of food in the body is an example of ____________ reaction
Anit [1.1K]

Answer:

A Chemical Reaction

Explanation:

There are 2 kinds of reactions physical and chemical. The digestion process often involves the use of digestive enzymes in your mouth and acid in your stomach.

4 0
4 years ago
1. What is the mass of a rock that has a density of 2.5 grams per cubic centimeter and a volume of 4.0
dangina [55]

Answer:

Explanation:

d =

m

V

m = d×V

V =

m

d

DENSITY

Density is defined as mass per unit volume.

d =

m

V

Example:

A brick of salt measuring 10.0 cm x 10.0 cm x 2.00 cm has a mass of 433 g. What is its density?

Step 1: Calculate the volume

V = lwh = 10.0 cm × 10.0 cm × 2.00 cm = 200 cm³

Step 2: Calculate the density

d =

m

V

=

433

g

200

c

m

³

= 2.16 g/cm³

MASS

d =

m

V

We can rearrange this to get the expression for the mass.

m = d×V

Example:

If 500 mL of a liquid has a density of 1.11 g/mL, what is its mass?

m = d×V = 500 mL ×

1.11

g

1

m

L

= 555 g

VOLUME

d =

m

V

We can rearrange this to get the expression for the volume.

V =

m

d

Example:

What is the volume of a bar of gold that has a mass of 14.83 kg. The density of gold is 19.32 g/cm³.

Step 1: Convert kilograms to grams.

14.83 kg ×

1000

g

1

k

g

= 14 830 g

Step 2: Calculate the volume.

V =

m

d

= 14 830 g ×

1

c

m

³

19.32

g

= 767.6 cm³

7 0
4 years ago
A student prepared several aqueous sodium chloride (NaCl) solutions to observe boiling point elevation at various molal concentr
NNADVOKAT [17]

The question is incomplete, the complete question is;

A student prepared several aqueous sodium chloride (NaCl) solutions to observe boiling point elevation at various molal concentrations, however, several errors were made throughout the procedure. Determine whether these errors would cause the observed boiling point to be increased or decreased relative to the expected boiling point, based on the procedure, or have no effect on the experimental results.

* The flask is washed with water but not thoroughly dried before preparing the solution

* The mass of NaCl used to prepare the solution is 5.400 grams instead of 4.400 grams

* Some of the prepared solution splashes out of the flask prior to observation of the boiling point

* When making the salt solution, 55.0 milliliters of water is added instead of 50.0 milliliters

Answer:

The flask is washed with water but not thoroughly dried before preparing the solution  - decrease

The mass of NaCl used to prepare the solution is 5.400 grams instead of 4.400 grams

-increase

Some of the prepared solution splashes out of the flask prior to observation of the boiling point- have no effect

When making the salt solution, 55.0 milliliters of water is added instead of 50.0 milliliters- decrease

Explanation:

If the flask is washed but not dried, then the solution will be further diluted than expected. As a result of this further dilution, the observed boiling point will be less than the expected boiling point because the boiling point depends on the solution's concentration.

If more solute is added than expected, the concentration of the solution is increased and the boiling point also increases above the expected boiling point due to the increase in amount of solute present.

If some of the solution splashes out of the flask prior to boiling, the boiling point is not affected because the concentration of the solution was not altered. The boiling point only changes when the concentration of the solution is changed.

If 55 ml of water is added instead of 50 ml, the solution is now more dilute than expected thereby reducing the concentration of the solution and the boiling point. Remember that, as the concentration of the solution decreases, the boiling point decreases accordingly.

8 0
3 years ago
Consider the reaction of ruthenium(III) iodide with carbon dioxide and silver. RuI3 (s) 5CO (g) 3Ag (s) Ru(CO)5 (s) 3AgI (s) Det
mixer [17]

Answer:

71.6 g of Ru(CO)₅ is the maximum mass that can be formed.

The limiting reactant is Ag

Explanation:

The reaction is:

RuI₃ (s) + 5CO (g) + 3Ag (s) → Ru(CO)₅ (s) + 3AgI (s)

Firstly we determine the moles of each reactant:

169 g . 1mol /481.77g = 0.351 moles of RuI₃

58g . 1mol /28g = 2.07 moles of CO

96.2g . 1mol/ 107.87g = 0.892 moles

Certainly, the excess reactant is CO, therefore, the limiting would be Ag or RuI₃.

3 moles of Ag react to 1 mol of RuI₃

Then 0.892 moles of Ag may react to (0.892 . 1) /3 = 0.297 moles

We have 0.351 moles of iodide and we need 0.297 moles, so this is an excess. In conclussion, Silver (Ag) is the limiting.

1 mol of RuI₃ react to 3 moles of Ag

Then, 0.351 moles of RuI₃ may react to (0.351 . 3) /1 = 1.053 moles

It's ok, because we do not have enough Ag. We only have 0.892 moles and we need 1.053.

5 moles of CO react to 3 moles of Ag

Then, 2.07 moles of CO may react to (2.07 . 3) /5 = 1.242 moles of Ag.

This calculate confirms the theory.

Now, we determine the maximum mass of Ru(CO)₅

3 moles of of Ag can produce 1 mol of Ru(CO)₅

Then 0.892 moles may produce (0.892 . 1) /3 = 0.297 moles

We convert moles to mass → 0.297 mol . 241.07g /mol = 71.6 g

8 0
3 years ago
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