__________ is a characteristic of an organism that allows it to survive better in an environment.

When 1.82 mole of HCL reacts with excess MnO2, how many moles of Cl2 form?

liubo4ka [24]

The balanced reaction is:

MnO2(s) + 4HCl(aq) → Cl2(g) + MnCl2(aq) + 2H2O(l)

We are given the amount of hydrochloric acid to be used for the reaction. This will be the starting point for the calculations.

1.82 mol HCl ( 1 mol Cl2 / 4 mol HCl) = 0.46 mol Cl2

Therefore, 0.46 mol of chlorine gas is produced for the reaction of hydrochloric acid and manganese oxide.

6 0

3 years ago

Q: What property of propane does Rodriguez change using the burner? Support your answer with evidence from the text.

UNO [17]

Answer:

The wind is starting to blow stronger, and when you’re riding in a basket under a hot air

balloon, just 400 feet above ground, that’s not necessarily a good thing. Keith Rodriguez looks

to the horizon and squints. He had planned to take off from Scioto Downs, a horse racetrack

south of Columbus, Ohio, fly a few miles north, and land his balloon in a barren cornfield next

to his pickup truck.

Then the wind changed. Instead of a light breeze from the south, now Rodriguez’s bright red

balloon is getting hit by stronger, colder winds from the west. He has plenty of propane fuel in

his tank—he probably could ride the wind halfway to Pennsylvania. But that would be

dangerous. Rodriguez’s choice of landing sites just became very limited. As the balloon

switches direction and floats east, everything below becomes a wide carpet of suburban

sprawl—big‐box stores, major highways, strip malls. Beyond the stores lie forests.

Explanation:

7 0

2 years ago

The reaction of 60.0 g of aluminum oxide with 30.0 g of carbon produced CO gas and 22.5 g of aluminum metal. What is the percent

natulia [17]

Answer:

The answer to your question is: 70.8%

Explanation:

Data

Al₂O₃ = 60 g

C = 30 g

CO = gas

Al = 22.5 g

MW Al₂O₃ = 102 g

MW C = 12 g

MW Al = 54 g

Reaction

Al₂O₃ + 3C ⇒ 3 CO + 2 Al

Limiting reactant

102 g of Al₂O₃ -------------- 54 g Al

60 g -------------- x

x = 31.8 g

36 g of C ------------------ 54 g of Al

30 g of C ------------------ x

x = 45 g of Al

Limiting reactant = Al₂O₃

Percent yield = $\frac{22.5}{31.8} x 100$

Percent yield = 70.75 %

5 0

2 years ago

Using the table below, what is the change in enthalpy for the following reaction? 3CO (g) + 2Fe2O3 (s) Imported Asset Fe(s) + 3C

zhuklara [117]

To solve this problem, we should recall that the change in enthalpy is calculated by subtracting the total enthalpy of the reactants from the total enthalpy of the products:

ΔH = Total H of products – Total H of reactants

You did not insert the table in this problem, therefore I will find other sources to find for the enthalpies of each compound.

ΔHf CO2 (g) = -393.5 kJ/mol

ΔHf CO (g) = -110.5 kJ/mol

ΔHf Fe2O3 (s) = -822.1 kJ/mol

ΔHf Fe(s) = 0.0 kJ/mol

Since the given enthalpies are still in kJ/mol, we have to multiply that with the number of moles in the formula. Therefore solving for ΔH:

ΔH = [3 mol ( − 393.5 kJ/mol) + 1 mol (0.0 kJ/mol)] − [3 mol ( − 110.5 kJ/mol) + 2 mol ( − 822.1 kJ/mol)]

ΔH = 795.2 kJ

3 0

2 years ago

What is the theoretical yield of vanadium, in moles, that can be produced by the reaction of 1.0 mole of V2O5 with 4.0 moles of

IRINA_888 [86]

Answer:

Theoretical yield of vanadium = 1.6 moles

Explanation:

Moles of $V_2O_5$ = 1.0 moles

Moles of $Ca$ = 4.0 moles

According to the given reaction:-

$V_2O_5_{(s)} + 5Ca_{(l)}\rightarrow 2V_{(l)} + 5CaO_{(s)}$

1 mole of $V_2O_5$ react with 5 moles of $Ca$

Moles of Ca available = 4.0 moles

Limiting reagent is the one which is present in small amount. Thus, Ca is limiting reagent. (4.0 < 5)

The formation of the product is governed by the limiting reagent. So,

5 moles of Ca on reaction forms 2 moles of V

1 mole of Ca on reaction for 2/5 mole of V

4.0 mole of Ca on reaction for $\frac{2}{5}\times 4$ mole of V

Moles of V = 1.6 moles

Theoretical yield of vanadium = 1.6 moles

4 0

3 years ago