You have nitrogen gas (N2) at a pressure of 0.5 atm, a volume of 25000 mL, and a temperature of 300 K, how many grams of nitroge

15mL because if it started as 20 mL and went to 35mL when u put in pyrite, the pyrite takes up 15mL

Enthalpy is a thermodynamic quantity that describes the heat content of a system, that can not be measured directly. That's why we measure change in enthaply, measured in the units joules. The statement that e<span>nthalpy change depends on the rate at which a substance is heated or cooled is false. Enthalpy change depends only on the following factors:
-</span><span>physical state of reactants and products
- quantity of reactants</span><span>
- allotropic modifications
- temperature and pressure</span><span>
</span>

Answer:

Explanation:

35.453 u

Answer:

72g

Explanation:

The balanced equation of the reaction between Magnesium (Mg) and hydrochloric acid (HCl):

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Based on this equation, 1 mole of Mg needs to react with 2 moles of HCl.

Hence, 6 moles of HCl will need to react with 6 × ½ = 3 moles of Mg.

Using the formula; mole = mass/molar mass

Molar mass of Mg = 24g/mol

3 = mass/24

mass = 72g.

Answer:

a. 510.6 g of C₂H₅OH are produced from 1kg of glucose

b. 171.1 g of glucose are required

Explanation:

Chemist reaction is this:

C₆H₁₂O₆  →  2C₂H₅OH(l) + 2CO₂(g)

So 1 mol of glucose can produce 1 mol of ethyl alcohol.

First of all, we should convert the mass to g, afterwards to moles

1 kg . 1000 g/ 1kg = 1000 g . 1 mol/180 g = 5.55 moles

Then we can think, this rule of three

1 mol of glucose can produce 2 moles of ethyl alcohol

Then 5.55 moles of glucose may produce the double of moles of C₂H₅OH

(5.55 .2)/1 = 11.1 moles.

Let's convert the moles to mass → 11.1 mol . 46g /1mol = 510.6 g

b. Let's determine the liters of ethyl alcohol we need.

1 gasohol is 10 mL C₂H₅OH / 90 ml of gasoline. We should make a rule of three.

In 90 mL of gasoline we have 10 mL of C₂H₅OH

In 1000 mL (1L) we would have (1000 . 10)/ 90 = 111.1 mL

Now we have to determine the mass of C₂H₅OH that is contained in the volume we have calculated. We must use the density.

Density = Mass /Volume

0.79 g/mL = Mass / 111.1 mL

0.79 g/mL . 111.1 mL = 87.7 g

Now, we convert the mass to moles → 87.7 g . 1mol/ 46g = 1.91 mol

Ratio is 2:1 so 2 moles of C₂H₅OH are produced by 1 mol of glucose

Therefore 1.91 mol would be produced by (1.91 .1)/2 = 0.954 moles

Finally we convert the moles of glucose to mass:

0.954 mol . 180 g/ 1mol = 171.7 grams.