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3 years ago

How many grams of oxygen are required to react completely with 3.6L of hydrogen at

1 answer:

7 0

the number of moles of oxygen required are 0.08 mol. The volume of oxygen that is required to react can be calculated by the formula shown below. Substitute the values in equation (II). Hence, the volume of oxygen required to react with 3.6 L hydrogen is 1.8L . I hope this helps if not I’m sorry

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1 i i r 1 X 1 I A mm mm nu. Inchon-Mvim u m an

Setler [38]

The answer is that the pilot was tired of life and committed suicide with hundreds of passengers.

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3 years ago

What kind of non-flowering plant produces seeds instead of spores?

Pie

Answer:

The answer to your question is Ferns

Explanation:

Ferns are called nonflowering plants and produce spores instead of seeds.

6 0

3 years ago

Consider the following scenario

GarryVolchara [31]

Answer:

See explaination

Explanation:

Going by the clues that it is between Silver Flouride (AgF) and Sodium Fluoride (NaF) and since it is an aqueous solution , the 1 liter bottle is likely to be Sodium Chloride( NaCl). Going by the reaction,

AgF + NaCl= AgCl + NaF

Here, the color of AgCl is white, hence the solution cannot be AgCl.

Determination of NaCl

Determination of NaCl can be done by Mohr's Method or Volhard's method. But results in Volhard's method are more accurate . Its uses the method of back titration with Potassium Thiocynate which forms a AgCl precipitate . Prior to titration,excess AgNO3 ( The problem also has a clue that excess reagents are present in the lab ) is added to the NaCl solution so that all the Cl- ions react with Ag+. Fe3+ is then added as an indicator and the solution is titrated with KSCN to form a silver thiocyannite precipitate (AgSCN). Once all the silver has reacted, a slight excess of SCN- reacts with Fe3+ to form Fe(SCN)3 dark red complex. The concentration of Cl- is determined by subtracting the titer findings of Ag+ ions that reacted to form AgSCN from the Ag NO3 moles added to the solution. This is used because pH of the solution is acidic. If the pH of solution is basic, Mohr's method is used.

Reactions

Ag+ (aq)+ Cl-(aq) = AgCl(aq)

Ag+(aq) + SCN-(aq) = AgSCN(aq)

Fe3+(aq) + SCN-(aq) = [FeSCN]2- (aq)

7 0

3 years ago

What is the limiting reagent when a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen?

UNO [17]

Answer:

Ammonia is limiting reactant

Amount of oxygen left = 0.035 mol

Explanation:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂ → 2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

NH₃ : N₂

4 : 2

0.12 : 2/4×0.12 = 0.06

NH₃ : H₂O

4 : 6

0.12 : 6/4×0.12 = 0.18

O₂ : N₂

3 : 2

0.125 : 2/3×0.125 = 0.08

O₂ : H₂O

3 : 6

0.125 : 6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

NH₃ : O₂

4 : 3

0.12 : 3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left = 0.125 - 0.09 = 0.035 mol

3 0

2 years ago

A solution of barium nitrate has 61.2g of barium nitrate in 1 liter of solution. How many mg of barium are there in 7.5 quarts

Nuetrik [128]

Answer:

220.44g Ba²⁺ ions in solution

Explanation:

Given parameters:

Mass of barium nitrate = 61.2g

Volume of solution = 1 liter

Unkown:

Mass of barium in 7.5quarts of solution?

Solution

We must first convert quarts to its liter equivalence:

1 quarts = 0.95 liter

7.5 quarts = 0.95 x 7.5; 7.125liter

Now, let us find the mass of barium nitrate in a solution of 7.125liter:

Given:

1 liter of solution contains 61.2g of barium nitrate:

7.125 liter will contain 7.125 x 61.2 = 436.05g of barium nitrate.

The formula of the compound is Ba(NO₃)₂:

In solution we have Ba²⁺ + NO₃⁻

Ba(NO₃)₂ → Ba²⁺ + 2NO₃⁻

Number of moles of Ba(NO₃)₂ = $\frac{mass of Ba(NO₃)₂}{Molar mass of Ba(NO₃)₂}$

Molar mass of Ba(NO₃)₂ = 137 + 2[14 + 3(16)] = 271g/mol

Number of moles of Ba(NO₃)₂ = $\frac{436.05}{271}$ = 1.609mole

1 mole of Ba(NO₃)₂ will produce 1 mole of Ba²⁺ ions in solution

therefore, 1.609mole of Ba(NO₃)₂ will also yield 1.609mole of Ba²⁺ ions in solution

Mass of Ba²⁺ ions = Number of moles of Ba²⁺ ions x molar mass of Ba²⁺ ions

Mass of Ba²⁺ ions = 1.609 x 137 = 220.44g

3 0

3 years ago