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OLEGan [10]
3 years ago
8

What are the procedures for performing an acid property test

Chemistry
1 answer:
Bond [772]3 years ago
8 0

Answer:

In the context of transaction processing, the acronym ACID refers to the four key properties of a transaction: atomicity, consistency, isolation, and durability. Atomicity. All changes to data are performed as if they are a single operation

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Which best describes nuclear fission? A nucleus spontaneously splits and absorbs energy. Two nuclei spontaneously combine and ab
Maurinko [17]

Answer:

A nucleus collides with a neutron and splits, releasing energy.

Explanation:

5 0
3 years ago
Read 2 more answers
1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to
Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant,  NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

MM:        17.03    32.00     30.01

              4NH₃  +  5O₂ ⟶ 4NO + 6H₂O

Mass/g:    1.5        1.85

2. Calculate the moles of each reactant  

\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

\text{Moles of NO} =  \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}

7. Calculate the mass of NH₃ reacted

\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0
3 years ago
What volume (L) will 3.56 mol NH3 occupy at STP. (1 point)
mart [117]

Answer:

B

Explanation:

I hope it helps you good luck

7 0
3 years ago
Someone help1!1!1!1!!!!!
alisha [4.7K]
The question is cut off in the picture
7 0
3 years ago
Liters of a 2.40 M
kotykmax [81]

Answer:

1.42 L

Explanation:

Step 1:

The following data were obtained from the question :

Molarity of KBr = 2.40 M

Mole of KBr = 3.40 moles

Volume of solution =?

Step 2:

Determination of the volume of the solution.

Molarity of solution is simply the mole of the solute per unit volume the of solution. It is given as :

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 3.4/2.4

Volume = 1.42 L

Therefore, the volume of the solution is 1.42 L

6 0
3 years ago
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