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2 years ago

What is the stoichiometric ratio between BaCl2 and NaCl

1 answer:

7 0

BaCl2+Na2SO4---->BaSO4+2NaCl There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent? "First convert grams into moles" 1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2 1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4 (7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2 "From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl" The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.

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A mixture of three noble gases has a total pressure of 1.25 atm. The individual pressures exerted by neon and argon are 0.68 atm

vlabodo [156]

Add the Pressure of neon and argon that is 0.68 +0.35= 1.03
Total pressure that is 1.25 -1.03=0.22 atm

5 0

1 year ago

Read 2 more answers

For scuba dives below 150 ft, helium is often used to replace nitrogen in the scuba tank. If 15.2 g of He(g) and 30.6 g of O2(g)

abruzzese [7]

Answer:

see explanation below

Explanation:

To do this exercise, we need to use the following expression:

P = nRT/V

This is the equation for an ideal gas. so, we have the temperature of 22 °C, R is the gas constant which is 0.082 L atm / mol K, V is the volume in this case, 5 L, and n is the moles, which we do not have, but we can calculate it.

For the case of the oxygen (AW = 16 g/mol):

n = 30.6 / 32 = 0.956 moles

For the case of helium (AW = 4 g/mol)_

n = 15.2 / 4 = 3.8 moles

Now that we have the moles, let's calculate the pressures:

P1 = 0.956 * 0.082 * 295 / 5

P1 = 4.63 atm

P2 = 3.8 * 0.082 * 295 / 5

P2 = 18.38 atm

Finally the total pressure:

Pt = 4.63 + 18.38

Pt = 23.01 atm

7 0

2 years ago

Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon

Pani-rosa [81]

Answer: The ratio of carbon in both the compounds is 1 : 2

Explanation:

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: $Cu_2O\text{ and }CuO$

For Sample 1:

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = $\frac{2.26}{2.26}=1$

For Oxygen = $\frac{4.55}{2.26}=2.01\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is $CO_2$

For Sample 2:

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = $\frac{3.57}{3.57}=1$

For Oxygen = $\frac{3.57}{3.57}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is $CO$

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = $\frac{1}{2}\times 1=\frac{1}{2}$

Now, taking the ratio of carbon atoms in both the samples, we get:

$C_1:C_2=\frac{1}{2}:1=1:2$

Hence, the ratio of carbon in both the compounds is 1 : 2

8 0

1 year ago

The flame produced by the burner of a gas (propane) grill is a pale blue color when enough air mixes with the propane (C3H8) to

adelina 88 [10]

Answer:

At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane

Explanation:

The combustion stoichiometry is as follows:

C₃H₈ + 5O₂ = 4 H₂O + 3CO₂ The molecular weights (g/mol) are:

MW 44 5x32 4x18 3x44

So each gram of propane is 1/44 = 0.02272 mol propane

and will need 5 x 0.02272 = 0.1136 mol oxygen

At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.

At the low pressure in the burner we can use the Ideal Gas Law

PV=nRT, or V = nRT/P

P = 1.1 x 101325 Pa = 111457 Pa

T = 195°C + 273 = 468 K

R = 8.314

and we calculated n = number of moles air = 0.54 mol

So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.

6 0

2 years ago

N a football game, two players tackle each other so hard that they both fly in opposite directions after they hit each other. �W

evablogger [386]

Answer:

Newton's Third Law

Explanation:

Newton's Third Law stipulates that for every action there is an equal and opposite reaction.

So when the two players are tackling they exert a force on each other.

If player 1 tackles (exerts a force) player 2, player 2 will exert an equal and opposite reaction on player 1 as stated in Newton's Third Law.

Therefore when they tackle each other so hard they both experience reaction forces so powerful that they fly in opposite directions.

Thus this is an example of the Newton's Third Law.

4 0

2 years ago