1. BaBr2(aq) + H2SO4(aq) →? please balance the equation and predict the products

What are some elements that can exist in room temperature?

SOVA2 [1]

Most chemical elements are solid at room temperature. Some elements exist as gases but only two elements, mercury (Hg) and bromine (Br) are liquids at room temperature (approx 25oC). Hope this helped you!

7 0

3 years ago

Limiting reactants would appreciate the help

Vanyuwa [196]

Answer:

Explanation:

The Limiting Reactant is that reactant which when consumed in a reaction stops the reaction. The other reactants will be in excess and typically considered non-reactive.

To identify the limiting reactant ...

- write and balance the reaction of interest. Express it in standard form. That is, standard form of a reaction is when the coefficients of the balanced equation are in their lowest whole number values. Also, remember that the standard equation is 'assumed' to be at STP conditions (0°C & 1atm).

- convert all given reactant values to moles

- divide each reactant mole value by the related coefficient of the the balanced standard equation. The smaller value is the limiting reactant. The remaining reactants will be in excess.

Your Problem:

Given: 3Ba + N₂ => Ba₃N₂

22.6g 4.2g ?

moles Ba => 22.6g/137.34g/mol = 0.165 mole Ba

moles N₂ => 4.2g/14.007g/mol= 0.150 mole N₂

Part A: Determining the Limited Reactant

Divide each mole value by respective coefficient ... smallest value is Limiting Reactant.

Barium => 0.165/3 = 0.055 <=> (Limiting Reactant)

Nitrogen => 0.15/1 = 0.15

Barium is the smaller result and is therefore the limiting reactant. This works for ALL limiting reactant type problems. However, be sure to use the mole values calculated first (Ba = 0.165mol & N₂ = 0.150mol) when doing ratio calculations.

Part B: Max (theoretical) amount of Ba₃N₂ produced:

Note: The product yield amounts are based upon the given 'moles' of limiting reactant, NOT the results of the 'divide by respective coefficient' step used to ID the limiting reactant.

3Ba + N₂ => Ba₃N₂ (3:1 rxn ratio for Ba:Ba₃N₂)

moles 0.165mole 0.150mole 1/3(0.165)mole = 0.055mole Ba₃N₂

= 0.055mol(440g/mol) Ba₃N₂

= 24.2 grams Ba₃N₂ (as based

upon Barium as Limiting Reactant)

Part C: Excess N₂ remaining after reaction stops:

From balanced standard reaction, the reaction ratio for Ba:N₂ is 3moles:1mole. That is, for the moles of Ba consumed, 1/3(moles of Ba) = moles of N₂ used.

moles of N₂ used = 1/3(0.165)mole = 0.055mole N₂ used

∴ the amount of N₂ remaining in excess = 0.150mole (given) - 0.055mole (used) = 0.095mole N₂ remaining in excess.

mass N₂ remaining = 0.095mole x 28g/mole = 2.66 grams N₂ remaining in excess.

5 0

3 years ago

THIS IS URGENT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Pani-rosa [81]

Answer:

1- 1.54 mol.

2- 271.9 kPa.

3- Yes, the tires will burst.

4- 235.67 kPa.

5- As, the temperature increased, the no. of molecules that has minimum kinetic energy increases as shown in image 1 that represents the Maxwell’s Distribution of Speeds of molecules. "Kindly, see the explanation and the attached images".

Explanation:

Q1- How many moles of nitrogen gas are in each tire?

To calculate the no. of moles of nitrogen gas in each tire, we can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the nitrogen gas (P = 247.0 kPa/101.325 = 2.44 atm),

V is the volume of the nitrogen gas (V = 15.2 L),

n is the no. of moles of the nitrogen gas (n = ??? mole),

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the nitrogen gas (T = 21°C + 273 = 294 K).

∴ n = PV/RT = (2.44 atm)(15.2 L)/(0.082 L/atm/mol.K)(294.0 K) = 1.54 mol.

Q2: What would the maximum tire pressure be at 50 degrees C?

Now, the temperature is raised to be 50°C (T = 50°C + 273 = 323 K).

The pressure can be calculated using the general gas law: PV = nRT.

∴ P = nRT/V = (1.54 atm)(0.082 L/atm/mol.K)(323.0 K)/(15.2 L) = 2.68 atm = 271.9 kPa.

Q3: Will the tires burst in Moses Lake? Explain.

Yes, the tires will burst because the internal pressure be 271.9 kPa that exceeds 270 kPa, the pressure above which the tires will burst.

Q4: If you must let nitrogen gas out of the tire before you go, to what pressure must you reduce the tires before you start your trip? (Assume no significant change in tire volume.)

To get the pressure that we must begin with:

Firstly, we should calculate the no. of moles at:

T = 55°C + 273 = 328 K,

Pressure = 270 kPa (the pressure above which the tires will burst). (P =270 kPa/101.325 = 2.66 atm).

V = 15.2 L, as there is no significant change in tire volume.

∴ n = PV/RT = (2.66 atm)(15.2 L)/(0.082 L.atm/mol.K)(328 K) = 1.5 mol.

1.5562 moles of N₂ in the tires will give a pressure of 270 kPa at 55°C, so this is the minimum moles of N₂ that will make the tires burst.

Now, we can enter this number of moles into the original starting conditions to tell us what pressure the tires will be at if we start with this number of moles of N₂.

P = ???

V = 15.6 L.

n = 1.5 mol

T = 21°C + 273 = 294.0 K

R = 0.0821 L.atm/mol.K.

∴ P = nRT/V = (1.5 mol x 0.082 x 294.0 K) / (15.6 L) = 2.2325 atm = 235.67 kPa.

So, the starting pressure needs to be 235.67 kPa or just under in order for the tires not to burst.

Q5: Create a drawing of the tire and show a molecular view of the air molecules in the tire at 247 kpa vs the molecular view of the air molecules after the tires have been heated. Be mindful of the number of molecules that you use in your drawing in the before and after scenarios. Use a caption to describe the average kinetic energy of the molecules in both scenarios.

As, the temperature increased, the no. of molecules that has minimum kinetic energy increases as shown in “image 1” that represents the Maxwell’s Distribution of Speeds of molecules.

The no. of molecules that possess a critical K.E. of molecules increases due to increasing the temperature activate the motion of molecules with high velocity as

(K.E. = 3RT/2), K.E. directly proportional to the temperature of the molecules (see image 2).

Also, the average speed of molecules increases as the K.E of the molecules increases (see image 3).

3 0

3 years ago

A student dissolves of glucose in of a solvent with a density of . The student notices that the volume of the solvent does not c

nikitadnepr [17]

Answer:

0.052 M

0.059 m

Explanation:

There is some missing info. I think this is the complete question.

A student dissolves 4.6 g of glucose in 500 mL of a solvent with a density of 0.87 g/mL. The student notices that the volume of the solvent does not change when the glucose dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits.

Step 1: Calculate the moles of glucose (solute)

The molar mass of glucose is 180.16 g/mol.

4.6 g × 1 mol/180.16 g = 0.026 mol

Step 2: Calculate the molarity of the solution

0.026 moles of glucose are dissolved in 500 mL (0.500 L) of solution. We will use the definition of molarity.

M = moles of solute / liters of solution

M = 0.026 mol / 0.500 L = 0.052 M

Step 3: Calculate the mass corresponding to 500 mL of the solvent

The solvent has a density of 0.87 g/mL.

500 mL × 0.87 g/mL = 435 g = 0.44 kg

Step 4: Calculate the molality of the solution

We will use the definition of molality.

m = moles of solute / kilograms of solvent

m = 0.026 mol / 0.44 kg = 0.059 m

4 0

3 years ago

When aluminium and chlorine compared , which has smaller atomic size ? Why ?

Zielflug [23.3K]

Chlorine has a smaller atomic size.
Explanation: As you move towards right of the periodic table, the atomic size decreases. This is because the number of protons is increasing towards the right of the periodic table, which applies a greater inward force on the electrons. that is why the elements on the right of the periodic table have a smaller atomic size when compared to the elements on the left. Since chlorine is on the right side of aluminium, it has a smaller atomic size.

7 0

2 years ago

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