What is the concentration of OH- in pure water at 298k

How many atoms are in 10.1 g Ne

LekaFEV [45]

Answer:

3.01 × 10²³ atoms Ne

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Tables
Moles

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

Identify

[Given] 10.1 g Ne

[Solve] atoms Ne

Step 2: Identify Conversions

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

[PT] Molar Mass of Ne: 20.18 g/mol\

Step 3: Convert

[DA] Set up: $\displaystyle 10.1 \ g \ Ne(\frac{1 \ mol \ Ne}{20.18 \ g \ Ne})(\frac{6.022 \cdot 10^{23} \ atoms \ Ne}{1 \ mol \ Ne})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 3.01398 \cdot 10^{23} \ atoms \ Ne$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

3.01398 × 10²³ atoms Ne ≈ 3.01 × 10²³ atoms Ne

4 0

3 years ago

Atoms that have a positive or negative electrical charge are called ________.

hjlf

Explanation:

They are called or known as cations

4 0

2 years ago

Why is it better to conduct an experiment more than once?

lilavasa [31]

The first reason to repeat experiments is simply to verify results. Different science disciplines have different criteria for determining what good results are. Biological assays, for example must be done in at least triplicate to generate acceptable data. Science is built on the assumption that published experimental protocols are repeatable.

2) The next reason to repeat experiments is to develop skills necessary to extend established methods and develop new experiments. “Practice make perfect” is true for the concert hall and the chemical laboratory.

3) Refining experimental observations is another reason to repeat. Maybe you did not follow the progress of the reaction like you should have.

4) Another reason to repeat experiments is to study and/or improve them in way. In the synthetic chemistry laboratory, for example, there is always a desire to improve the yield of a synthetic step. Will certain changes in the experimental conditions lead to a better yield? The only way to find out is to try it! The scientific method informs us that it is best to only make one change at a time.

5) The final reason to repeat an extraction, chromatographic or synthetic protocol is to produce more of your target substance. This is sometimes referred to scale-up.

8 0

2 years ago

Ammonia, NH 3 , may react with oxygen to form nitrogen gas and water. 4 NH 3 ( aq ) + 3 O 2 ( g ) ⟶ 2 N 2 ( g ) + 6 H 2 O ( l )

Alex Ar [27]

Answer:

The limiting reactant is NH₃

0.0186moles of N₂ are the one produced by the limiting reactant

0.020 moles of N₂ are the one produced by the reactant in excess

Explanation:

This is the reaction

4NH₃ + 3O₂ → 2N₂ + 6H₂O

We should calculate the moles of each reactant

Mass / Molar mass = Moles

3.55 g / 17g/m = 0.208 moles NH₃

5.33 g / 32g/m = 0.166 moles O₂

4 moles of ammonia react with 3 moles of oxygen

0.208 moles of ammonia react with (0.208 .3)/4 = 0.156 moles O₂

We have 0.166 moles of O₂ and we need 0.156 moles, so O₂ is the reactant in excess.

3 moles of O₂ react with 4 moles of NH₃

0.166 moles of O₂ react with (0.166 . 4)/ 3 = 0.221 moles

We have 0.208 moles NH₃ and we need 0.221, so NH₃ is the limiting reactant.

To know the moles of N₂, let's apply the Ideal Gas Law

P.V =n.R.T

1atm . 0.450L = n . 0.082 . 295K

0.450 / (0.082 .295) = 0.0186 moles

If we have 100 % yield reaction:

4 moles NH₃ make 2 moles N₂

0.208 moles NH₃ make (0.208 .2)/4 = 0.104 moles

So the % yield reaction is.

0.104 moles ___ 100%

0.0186 moles ___ 17.9%

0.0186moles of N₂ are the one produced by the limiting reactant.

3 moles of O₂ produce 2 moles N₂

0.166 moles O₂ produce (0.166 .2)/3 = 0.111 moles

Now, we apply the yield.

100% ____ 0.111 moles

17.9% = 0.020 moles

8 0

2 years ago

Read 2 more answers

The specific heat of aluminum is 0.214 cal/g.oC. Determine the energy, in calories, necessary to raise the temperature of a 55.5

Natasha2012 [34]

For this problem, we use the formula for sensible heat which is written below:

Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference

Q = (55.5 g)(0.214 cal/g·°C)(48.6°C- 23°C)
Q = 304.05 cal

4 0

3 years ago